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      Class 10 Maths

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      • Class 10
      • Class 10 Maths
      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4

        Q.1    Check whether the following are quadratic equations :
                  (i) (x+1)2=2(x−3)
                  (ii) x2−2x=(−2)(3−x)
                  (iii) (x−2)(x+1)=(x−1)(x+3)
                  (iv) (x−3)(2x+1)=x(x+5)
                  (v) (2x−1)(x−3)=(x+5)(x−1)
                  (vi) x2+3x+1=(x−2)2
                  (vii) (x+2)3=2x(x2−1)
                  (viii) x3−4x2−x+1=(x−2)3
        Sol.    (i) We have (x+1)2=2(x−3)
        ⇒ x2+2x+1=2x−6
        ⇒ x2+2x+1−2x+6=0
        ⇒ x2+7=0
        Clearly, x2+7 is a quadratic polynomial. So the given equation is a quadratic equation.

        (ii) We have , x2−2x=(−2)(3−x)
        ⇒ x2−2x+2(3−x)=0
        ⇒ x2−2x+6−2x=0
        ⇒ x2−4x+6=0
        Clearly, x2−4x+6 is a quadratic polynomial. SO, the given equation is a quadratic equation.

        (iii) We have , (x−2)(x+1)=(x−1)(x+3)
        ⇒ x2−x−2=x2+2x−3
        ⇒ x2−x−2−x2−2x+3=0
        ⇒ −3x+1=0
        Clearly, −3x+1 is linear polynomial  So the given equation is not a quadratic equation.

        (iv) We have, (x−3)(2x+1)=x(x+5)
        ⇒ x(2x+1)−3(2x+1)−x(x+5)=0
        ⇒ 2x2+x−6x−3−x2−5x=0
        ⇒ x2−10x−3=0
        Clearly, x2−10x−3 is a quadratic polynomial. So, the given equation is a quadratic equation.

        (v) We have, (2x−1)(x−3)=(x+5)(x−1)
        ⇒ (2x−1)(x−3)−(x+5)(x−1)=0
        ⇒ 2x(x−3)−1(x−3)−x(x−1)−5(x−1)=0
        ⇒ 2x2−6x−x+3−x2+x−5x+5=0
        ⇒ x2−11x+8=0
        Clearly, x2−11x+8 is a quadratic polynomial. So, the given equation is a quadratic equation.

        (vi) We have x2+3x+1=(x−2)2
        ⇒ x2+3x+1−(x−2)2=0
        ⇒ x2+3x+1−(x2−4x+4)=0
        ⇒ x2+3x+1−x2+4x−4=0
        ⇒ 7x−3=0
        Clearly, 7x−3 is a linear polynomial. So, the given equation is not a quadratic equation.

        (vii) We have, (x+2)3=2x(x2−1)
        ⇒ x3+3x2(2)+3x(2)2+(2)3=2x3−2x
        ⇒ x3+6x2+12x+8−2x3+2x=0
        ⇒ −x3+6x2+14x+8=0
        Clearly, −x3+6x2+14x+8 being a polynomial of degree 3, is not a quadratic polynomial. So the given equation is not a quadratic equation.

        (viii) We have, x3−4x2−x+1
        =x3+3x2(−2)+3x(−2)2+(−2)3
        ⇒ x3−4x2−x+1=x3−6x2+12x−8
        ⇒ x3−4x2−x+1−x3+6x2−12x+8=0
        ⇒ 2x2−13x+9=0
        Clearly, 2x2−13x+9 is a quadratic polynomial. So, the given equation is a quadratic equation.


        Q.2       Represent the following situation in the form of quadratic equations.
                      (i) The area of a rectangular plot is 528m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
                      (ii) The product of two consecutive positive integers is 306. We need to find the integers.
                      (iii) Rohan’s mother is 26 year older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
                      (iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

        Sol.      (i) Let the length and breadth of the rectangular plot be 2x + 1 metres and x metres respectively. It is given that its area = 528m2.
        Since (2x+1)×x=528
        ⇒ 2x2+x=528
        ⇒ 2x2+x−528=0,
        Which is the required quadratic equation satisfying  the given conditions.

        (ii) Let two consecutive integers be x and x + 1 such that their product = 306.
        ⇒ x (x + 1) = 306
        ⇒ x2+x−306=0
        Which is the required quadratic equation satisfying  the given conditions.

         

        (iii) Let Rohan ‘s present age be x years. Then,
        His mother’s age = (x + 26) years.
        After 3 years, their respective ages are (x + 3) years and (x + 29) years. … (1)
        It is given that the product of, ages mentioned at (1) is 360
        i.e. (x + 3)(x + 29) = 360
        ⇒ x2+32x+87=360
        ⇒ x2+32x+87−360=0
        ⇒ x2+32x−273=0
        Therefore, the age of Rohan satisfies the quadratic equation x2+32x−273=0.

        (iv) Let u km/hr be the speed of the train.
        Then, time taken to cover 480 km =480uhours
        Time taken to cover 480 km when the speed  is decreased. by 8 km/hr
        =480u−8hours
        It is given that the time to cover 480 km is increased by 3 hours.
        Therefore, 480u−8−480u=3
        ⇒ 480u−480(u−8)=3u(u−8)
        ⇒ 160u−160u+1280=u2−8u
        ⇒ u2−8u−1280=0
        Therefore, the speed of the train satisfies the quadratic equation u2−8u−1280=0

         

         

        Q.1     Find the roots of the following quadratic equations by factorization :
                   (i) x2−3x−10=0
                   (ii) 2x2+x−6=0
                   (iii) 2–√x2+7x+52–√=0
                   (iv) 2x2−x+18=0
                   (v) 100x2−20x+1=0
        Sol.     (i) We have, x2−3x−10=0
        ⇒ x2−5x+2x−10=0
        ⇒ x(x−5)+2(x−5)=0
        ⇒ (x−5)(x+2)=0
        ⇒ x−5=0orx+2=0
        ⇒ x=5orx=−2
        Thus , x = 5 and x = – 2 are two roots of the equation x2−3x−10=0.

        (ii) We have, 2x2+x−6=0
        ⇒ 2x2+4x−3x−6=0
        ⇒ 2x(x+2)−3(x+2)=0
        ⇒ (x+2)(2x−3)=0
        ⇒ x+2=0or2x−3=0
        ⇒ x=−2orx=32
        Thus x=−2orx=32 are two roots of the equation 2x2+x−6=0

         

        (iii) We have, 2–√x2+7x+52–√=0
        ⇒ 2–√x2+2x+5x+52–√=0
        ⇒ 2–√x(x+2–√)+5(x+2–√)=0
        ⇒ (x+2–√)(2–√x+5)=0
        ⇒ x+2–√=0
        or 2–√x+5=0
        ⇒ x=−2–√
        or x=−52√=−52√2
        Thus, x=−2–√andx=−52√2 are two roots of the equation 2–√x2+7x+52–√=0.

        (iv) We have, 2x2−x+18=0
        ⇒ 16x2−8x+1=0
        ⇒ 16x2−4x−4x+1=0
        ⇒ 4x(4x−1)−1(4x−1)=0
        ⇒ (4x−1)(4x−1)=0
        ⇒ 4x−1=0
        or 4x−1=0
        ⇒ x=14orx=14
        Thus x=14orx=14 are two roots of the equation
        2x2−x+18=0.

        (v) We have , 100x2−20x+1=0
        ⇒ 100x2−10x−10x+1=0
        ⇒ 10x(10x−1)−1(10x−1)
        ⇒ (10x−1)(10x−1)=0
        ⇒ 10x−1=0
        or 10x−1=0
        ⇒ x=110orx=110
        Thus , x=110orx=110 are two roots of the equation 100x2−20x+1=0


        Q.2      Solve the problems given in Example 1 i.e., to solve
                    (i) x2−45x+324=0
                    (ii) x2−55x+750=0
                    using factorisation method.
        Sol.     (i) We have , x2−45x+324=0
        ⇒ x2−9x−36x+324=0
        ⇒ x(x−9)−36(x−9)=0
        ⇒ (x−9)(x−36)=0
        ⇒ x−9=0orx−36=0
        ⇒ x=9orx=36
        Thus x = 9 and x = 36 are two roots of the equation x2−45x+324=0.

        (ii) We have, x2−55x+750=0
        ⇒ x2−30x−25x+750=0
        ⇒ x(x−30)−25(x−30)=0
        ⇒ (x−30)(x−25)=0
        ⇒ x−30=0orx−25=0
        ⇒ x=30orx=25
        Thus, x = 30 and x = 25 are two roots of the equation x2−55x+750=0.


        Q.3     Find two numbers whose sum is 27 and product is 182.

        Sol.     Let the required numbers be x and 27 – x. Then x(27−x)=182
        ⇒ 27x−x2=182
        ⇒ x2−27x+182=0
        ⇒ x2−13x−14x+182=0
        ⇒ x(x−13)−14(x−13)=0
        ⇒ (x−13)(x−14)=0
        ⇒ x−13=0orx−14=0
        ⇒ x=13orx=14
        Hence, the two numbers are 13 and 14.


        Q.4      Find two consecutive positive integers, sum of whose squares is 365.
        Sol.      Let the two consecutive positive integers be x and x + 1.
        Then , x2+(x+1)2=365
        ⇒ x2+x2+2x+1=365
        ⇒ 2x2+2x−364=0
        ⇒ x2+x−182=0
        ⇒ x2+14x−13x−182=0
        ⇒ x(x+14)−13(x+14)=0
        ⇒ (x+14)(x−13)=0
        ⇒ x+14=0orx−13=0
        ⇒ x=−14orx=13
        Since x, being a positive integer, cannot be negative.
        Therefore, x = 13.
        Hence, the two consecutive positive integers are 13 and 14.


        Q.5      The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
        Sol.      Let the base of he right angled  triangle be x cm.
        Then, its height is (x – 7) cm.
        It given that the hypotenuse = 13
        ⇒ x2+(x−7)2−−−−−−−−−−−√=13
        ⇒ x2+(x2−14x+49)=169
        ⇒ 2x2−14x−120=0
        ⇒ x2−7x−60=0
        ⇒ x2−12x+5x−60=0
        ⇒ x(x−12)+5(x−12)=0
        ⇒ (x−12)(x+5)=0
        ⇒ x−12=0orx+5=0
        ⇒ x=12orx=−5
        ⇒ x=12
        [Since, side of a triangle can never be negative]
        Therefore, Length of the base = 12 cm and , Length of the height = (12 – 7) cm = 5 cm.


        Q.6     A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced On that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
        Sol.     Let the number of articles produced in a day = x
        Then, cost of  production of each article = Rs (2 x + 3)
        It is given that the total cost of production = Rs 90
        Therefore, x × (2 x + 3) = 90
        ⇒ 2x2+3x−90=0
        ⇒ 2x2−12x+15x−90=0
        ⇒ 2x(x−6)+15(x−6)=0
        ⇒ (x−6)(2x+15)=0
        ⇒ x−6=0or2x+15=0
        ⇒ x=6orx=−152
        ⇒ x=6
        [Since, the number of articles produced cannot be negative]
        Cost of each article = Rs (2 × 6 + 3) = Rs 15
        Hence, the number of articles produced are 6 and the cost of each article is Rs 15.

         

        Q.1      Find the roots of the following quadratic equations, if they exist by the method of completing the square:
                    (i) 2x2−7x+3=0
                    (ii) 2x2+x−4=0
                    (iii) 4x2+43–√x+3=0
                    (iv) 2x2+x+4=0
        Sol.      (i) The equation 2x2−7x+3=0 is the same as x2−72x+32=0
        Now, x2−72x+32
        =(x−74)2−(74)2+32
        =(x−74)2−4916+32
        =(x−74)2−2516
        Therefore, 2x2−7x+3=0
        ⇒ (x−74)2−2516=0
        ⇒ (x−74)2=2516
        ⇒ x−74=±54
        ⇒ x=74±54
        ⇒ x=74+54=124=3
        ⇒ x=74−54=24=12
        Therefore, The roots of the given equation are 3 and 12.

        (ii) We have, 2x2+x−4=0
        ⇒ x2+x2−2=0
        ⇒ (x+14)2−(14)2−2=0
        ⇒ (x+14)2−116−2=0
        ⇒ (x+14)2−3316=0
        ⇒ (x+14)2=3316
        ⇒ x+14=±33√4
        ⇒ x=−14±33√4
        ⇒ x=−1+33√4
        ⇒ x=−1−33√4
        Therefore, The roots of the given equation are −1−33√4 and −1+33√4
        (iii) We have, 4x2+43–√x+3=0
        ⇒ (2x)2+2×(2x)×3–√+(3–√)2−(3–√)2+3=0
        ⇒ (2x+3–√)2−3+3=0
        ⇒ (2x+3–√)2=0
        ⇒ x=−3√2
        ⇒ x=−3√2
        Therefore, The roots of the given equation are −3√2 and −3√2.

        (iv) We have, 2x2+x+4=0
        ⇒ x2+12x+2=0
        ⇒ (x+14)2−116+2=0
        ⇒ (x+14)2+3116=0
        ⇒ (x+14)2=−3116<0
        But (x+14)2 cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.


        Q.2      Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
        Sol.      (i) The given equation is 2x2−7x+3=0
        Here a = 2, b = – 7 and c = 3.
        Therefore, D=b2−4ac=(−7)2−4×2×3 = 49−24=25>0
        So, the given equation has real roots given by
        x=−b±D√2a
        =−(−7)±25√2×2
        =7±54
        =124or24=3or12

        (ii) The given equation is 2x2+x−4=0
        Here, a = 2, b = 1 and c = – 4
        Therefore, D=b2−4ac=(1)2−4×2×−4=1+32=33>0
        So, the given equation has real roots given by
        x=−b±D√2a=−1±33√2×2 =−1±33√4

        (iii) The given equation is 4x2+43–√x+3=0
        Here a = 4, b=43–√andc=3
        Therefore, D=b2−4ac=(43–√)2−4×4×3=48−48=0
        So, the given equation has real equal roots given by
        x=−b±D√2a=−b2a =−43√2×4=−3√2
        (iv) The given equation is 2x2+x+4=0
        Here, a = 2, b = 1 and c = 4
        Therefore, D=b2−4ac=(1)2−4×2×4=1−32=−31<0
        So, the given equation has no real roots.
        I prefer to use quadratic formula method as it is a straight forward method.


        Q.3      Find the roots of the following equations :
                    (i) x−1x=3,x≠0
                    (ii) 1x+4−1x−7=1130,x≠−4,7
        Sol.       (i) The given equation is x−1x=3,x≠0
        ⇒ x2−3x−1=0
        Here , a = 1, b = – 3 and c = – 1
        Therefore, D=b2−4ac=(−3)2−4(1)(−1)=9+4=13>0
        So, the given equation has real roots given by
        x=−b±D√2a=−(−3)±13√2×1 =3±13√2

        (ii) The given equation is 1x+4−1x−7=1130,x≠−4,7
        ⇒ (x−7)−(x+4)(x+4)(x−7)=1130
                     ⇒x−7−x−4(x+4)(x−7)=1130
        ⇒ −11(x+4)(x−7)=1130
                     ⇒−1x2−7x+4x−28=130
                     ⇒ −1x2−3x−28=130
                     ⇒ −30=x2−3x−28
        ⇒ x2−3x+2=0
                    ⇒x2−2x−x+2=0
                     ⇒ (x−1)(x−2)=0
                     ⇒x(x−2)−1(x−2)=0
        ⇒ x=1or2
        Thus x = 1 and x = 2 are the roots of the given equation.


        Q.4      The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 13. Find his present age.
        Sol.      Let Rehman’s present age be x – years.
        As per question, we have
        1x−3+1x+5=13
        ⇒ 3(x+5)+3(x−3)=(x−3)(x+5)
        ⇒ 3x+15+3x−9=x2+2x−15
        ⇒ x2−4x−21=0
        ⇒ x2−7x+3x−21=0
                    ⇒ (x−7)(x+3)=0
                    ⇒x(x−7)+3(x−7)=0
                    ⇒(x+3)(x−7)=0
        ⇒ x=7or−3
        Therefore, x=7 [Since, age can never be negative]
        Thus, Rehman’s present age is 7 years.


        Q.5      In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
        Sol.      Let Shefali’s marks in Mathematics be x. Then her marks in English will be (30 – x).
        As per condition of the problem :
                    (x+2)×[(30−x)−3]=210
        ⇒ (x+2)(27−x)=210
        ⇒ 27x−x2+54−2x=210
                    ⇒ x2−25x+156=0
                    ⇒ x2−12x−13x+156=0
                    ⇒ x(x−12)−13(x−12)=0
                    ⇒ (x−12)(x−13)=0
                    ⇒ x=12orx=13
                Therefore, Shafali’s marks in Mathematics and English are 12 and 18 respectively or in Mathematics and English are 13 and 17, respectively.


        Q.6      The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
        Sol.      Let in the rectangular field BC = x metres. Then AC = (x + 60) and AB = (x + 30) metres. By Pythagoras Theorem, we have

        13           AC2=BC2+AB2
                   ⇒ (x+60)2=x2+(x+30)2
                   ⇒ x2+120x+3600=x2+x2+60x+900
                   ⇒ x2−60x−2700=0
                   ⇒ x2−90x+30x−2700=0
                   ⇒ x(x−90)+30(x−900)=0
        ⇒ (x+30)(x−90)=0
        ⇒ x=−30orx=90
                   ⇒ x=90 [Since side of a rectangle can never be negative]
                   Hence, the sides of the field are 120 m and 90 m.


        Q.7      The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
        Sol.    Let the larger number be x. Then ,
                  Square of the smaller number = 8x
                  Also , square of the larger number = x2
                  It is given that the difference of the squares of the number is 180
                  Therefore, x2−8x=180
                  ⇒ x2−8x−180=0
                  ⇒ x2−18x+10x−180=0
                  ⇒ x(x−18)+10(x−18)=0
                  ⇒ (x−18)(x+10)=0
                  ⇒ x−18=0orx+10=0
                  ⇒ x=18orx=−10
                  Case 1 : When x = 18. In this case, we have
                  Square of the smaller number = 8x = 8 × 18 = 144
                  Therefore, Smaller number =±12
                  Thus, the number are 18, 12 or 18, – 12.
                  Case 2 : When x = – 10
                  In this case, we have
                  Square of the smaller number = 8x = 8 × – 10 = – 80
                  But , square of a number is always positive.
                  Therefore, x = – 10 is not possible.
                  Hence, the numbers are 18, 12 or 18 , – 12.


        Q.8      A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
        Sol.       Let x km/hr be the uniform speed of the train.
                    Then time taken to cover 360km=360xhours.
                    Time taken to cover 360 km when the speed is increased by 5km/hr=360x+5hours.
        It is given that the time to cover 360 km is reduced by 1 hour.
        Therefore, 360x−360x+5=1
                    ⇒ 360(x+5)−360x=x(x+5)
                    ⇒ 360x+1800−360x=x2+5x
                    ⇒ x2+5x−1800=0
                    ⇒ x2+45x−40x−1800=0
                    ⇒ x(x+45)−40(x+45)=0
                    ⇒ (x−40)(x+45)=0
        ⇒ x−40=0orx+45=0
        ⇒ x=40orx=−45
                    But x cannot be negative. Therefore, x = 40.
                    Hence, the original speed of the train is 40 km/hr.


        Q.9     Two water taps together can fill a tank in 938 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
        Sol.        Let the smaller tap takes x hours to fill the tank. So, the larger tap will take (x – 10) hours to fill the tank.
        Therefore, Portion of the tank filled by the larger tap in one hours =1x−10
        ⇒ Portion of the tank filled by the larger tap 938hours
        i.e., 758hours=1x−10×758
        Similarly, portion of the tank filld by the smaller tap in 758hours=1x×758
        Since it is given that the tank is filled in 758hours
        Therefore, 758(x−10)+758x=1
        ⇒ 1x−10+1x=875
        ⇒ x+x−10x(x−10)=875
        ⇒ 75(2x−10)=8x(x−10)
        ⇒ 150x−750=8x2−80x
        ⇒ 8x2−230x+750=0
        ⇒ 4x2−115x+375=0
                     ⇒ 4x2−100x−15x+375=0
                     ⇒ 4x(x−25)−15(x−25)=0
                     ⇒ (x−25)(4x−15)=0
                     ⇒ x−25=0or4x−15=0
                     ⇒ x=25or154
                     Therefore, x=25asx=154 is inadmissible
                     Hence, the larger tap fills the tank in 15 hours and the smaller tap takes 25 hours to fill the tank.


        Q.10      An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, ,find the average speed of the two trains.
        Sol.      Let the speed of the express train be x km/hr.
        Then, the speed of the passenger train is (x – 11) km/hr
        Distance to be covered by train = 132
        Time taken by express train =132xhours and, time taken by the passenger train =132x−11
        By the given condition, 132x−11−132x=1
        ⇒ 132x−132x+1452=x2−11x
        ⇒ x2−11x−1452=0
        ⇒ x2−44x+33x−1452=0
        ⇒ x(x−44)+33(x−44)=0
        ⇒ (x−44)(x+33)=0
        ⇒ x=44
        or x=−33
        But x cannot be negative. Therefore, x = 44
        Therefore, Speed of express train = 44 km/hr
        and, speed of passenger train = (44 – 11) km/hr = 33 km/hr.


        Q.11     Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
        Sol.       Let the sides of the squares be x and y metres (x > y).
        According to question :
                     x2+y2=468 … (1)
                     and 4x−4y=24
                     ⇒ x−y=6 … (2)
                     Putting x = y + 6 in (1), we get (y+6)2+y2=468
                    ⇒y2+12y+36+y2=468
        ⇒ 2y2+12y+36−468=0
                    ⇒2y2+12y−432=0
        ⇒ y2+6y−216=0
                    ⇒ y2+18y−12y−216=0
                    ⇒ y(y+18)−12(y+18)=0
                    ⇒ (y+18)(y−12)=0
                    ⇒ y=−18ory=12
                    But y cannot be negative. Therefore, y = 12
                    Therefore, x = y + 6 = 12 + 6 = 18
                    Therefore, The sides of the squares are 18 m and 12 m.

         

         

        Q.1      Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
                    (i) 2x2−3x+5=0
                    (ii) 3x2−43–√x+4=0
                    (iii) 2x2−6x+3=0
        Sol.       (i) The given equation is 2x2−3x+5=0
                     Here , a = 2, b = – 3 and c = 5
                     Therefore, D=b2−4ac=(−3)2−4×2×5=9−40=−31<0
                     So, the given equation has no real roots.

        (ii) The given equation is 3x2−43–√x+4=0
        Here a = 3, b=−43–√andc=4
        Therefore, D=b2−4ac=(−43–√)2−4×3×4=48−48=0
        So, the given equation has real equal roots, given by
        x=−b±D√2a=−(−43√)±02×2=3–√.

        (iii) The given equation is 2x2−6x+3=0
        Here, a = 2, b = – 6 and c = 3
        Therefore, D=b2−4ac=(−6)2−4×2×3=36−24=12>0
        So, the given equation has real roots , given by
        x=−b±D√2a=−(−6)±12√2×2
        =6±23√4=3±3√2


        Q.2      Find the values of k for each of the following quadratic equations, so that they have two equal roots.
                    (i) 2x2+kx+3=0
                    (ii) kx(x−2)+6=0
        Sol.      (i) The given equation is 2x2+kx+3=0
        Here, a = 2, b = k and c = 3
        Therefore, D=b2−4ac=k2−4×2×3=k2−24
        The given equation will have real and equal roots, if
        D = 0 ⇒ k2−24=0 ⇒ k=±24−−√=±26–√
        (ii) The given equation is kx (x – 2) + 6 = 0
        ⇒ kx2−2kx+6=0
        Here a = k , b = – 2k and c = 6
        Therefore, D=b2−4ac=(−2k)2−4×k×6=4k2−24k
        The given equation will have real and equal roots, if D = 0
        ⇒ 4k2−24k=0 ⇒ 4k(k−6)=0
        ⇒ k=0ork=6


        Q.3      Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2 ? If so, find its length and breadth.
        Sol.      Let 2x be the length and x be the breadth of a rectangular mango grove.
        Area = (2x) (x) = 800 [Given]
        ⇒ x2=400
        ⇒ x=20 [Since cannot  be negative]
        The value of x is real so design of grove is possible
        Its length = 40 m and breadth = 20 m.


        Q.4      Is the following situation possible ? If so, determine their present ages. The sum of the ages of two friend is 20 years. Four year ago, the porudct of their ages in years was 48.
        Sol.      Let age of one of the friends = x years
        Then, age of the other friend = 20 – x Four years ago,
                    Age of one of the friend = (x – 4) years
                    and age of the other friend = (20 – x – 4) years
                    = (16 – x) years
                    According to condition :
                    (x−4)(16−x)=48
                    ⇒ 16−x2−64+x=48
                    ⇒ x2−20+112=0
                    Here a = 1, b = – 20 and c = 112
                    Therefore, D=b2−4ac=(−20)2−4×1×112
                    =400−448=−48<0
                    So, the given equation has no real roots.
                    Thus , the given situation is not possible.


        Q.5      Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
        Sol.      Let length be x metres and breadth be y metres.
        Therefore, Perimeter = 80 m
        ⇒ 2(x+y)=80
        ⇒ x+y=40 …(1)
        Also, Area=400m2
        ⇒ xy=400
        ⇒ x(40−x)=400 [Using (1)]
        ⇒ 40x−x2=400
        ⇒ x2−40x+400=0
        Here a = 1, b = – 40 and c = 400
        Therefore,D=b2−4ac
        =(−40)2−4×1×400
                   =1600−1600=0
                   So, the given equation has equal real roots.
                   Therefore, Its length and breadth is given by
                   x2−40x+400=0
                   ⇒ (x−20)2=0
                   ⇒ x=20,20
                   Therefore, Length = 20 m
                   Breadth = 20 m
                   Therefore, Design is possible.

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