NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 – Quadratic Equations

Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 – Quadratic Equations. This Exercise contains 11 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 – Quadratic Equations

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NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 – Quadratic Equations

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x2 – 7x + 3 = 0

 Solution:  The quadratic equation is

2x2 – 7x + 3 = 0

On Dividing both side by 2,

x2 \cfrac { 7 }{ 2 } x + \cfrac { 3 }{ 2 } = 0

x2 – 2.\cfrac { 7 }{ 4 } .x + { \left( \cfrac { 7 }{ 4 } \right) }^{ 2 } { \left( \cfrac { 7 }{ 4 } \right) }^{ 2 } + \cfrac { 3 }{ 2 } =0

{ \left( x-\cfrac { 7 }{ 4 } \right) }^{ 2 }\cfrac { 49 }{ 16 } + \cfrac { 3 }{ 2 } =0

{ \left( x-\cfrac { 7 }{ 4 } \right) }^{ 2 } = \cfrac { 25 }{ 16 }

{ \left( x-\cfrac { 7 }{ 4 } \right) } = ±\cfrac { 5 }{ 4 }

Taking the positive value,

x – \cfrac { 7 }{ 4 } = \cfrac { 5 }{ 4 }

⇒ x = \cfrac { 7 }{ 4 } + \cfrac { 5 }{ 4 }  = \cfrac { 12 }{ 4 } = 3

And taking the negative value,

x – \cfrac { 7 }{ 4 } = –\cfrac { 5 }{ 4 }

⇒ x = \cfrac { 7 }{ 4 } \cfrac { 5 }{ 4 }  = \cfrac { 2 }{ 4 } = \cfrac { 1 }{ 2 }

Hence the roots are x = 3, \cfrac { 1 }{ 2 }

(ii) 2x2+ x – 4 = 0

Solution: On Dividing both sides by 2

x2 + \cfrac { x }{ 2 } – 2 = 0

x2+2.x.\cfrac { 1 }{ 4 }  +{ \left( \cfrac { 1 }{ 4 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 4 } \right) }^{ 2 } -2 = 0

{ \left( x+\cfrac { 1 }{ 4 } \right) }^{ 2 } – \cfrac { 1 }{ 16 } -2 = 0

{ \left( x+\cfrac { 1 }{ 4 } \right) }^{ 2 }\cfrac { 33 }{ 16 } = 0

{ \left( x+\cfrac { 1 }{ 4 } \right) }^{ 2 } = \cfrac { 33 }{ 16 }

x + \cfrac { 1 }{ 4 } = ±\cfrac { \sqrt { 33 } }{ 4 }

Taking the positive value,

x + \cfrac { 1 }{ 4 } = \cfrac { \sqrt { 33 } }{ 4 }

⇒ x = –\cfrac { 1 }{ 4 } + \cfrac { \sqrt { 33 } }{ 4 }

⇒ x = \cfrac { -1+\sqrt { 33 } }{ 4 }

And taking the negative sign,

x + \cfrac { 1 }{ 4 } = –\cfrac { \sqrt { 33 } }{ 4 }

⇒ x = –\cfrac { 1 }{ 4 } \cfrac { \sqrt { 33 } }{ 4 }

⇒ x = \cfrac { -1-\sqrt { 33 } }{ 4 }

Hence the required roots are \cfrac { -1+\sqrt { 33 } }{ 4 } and \cfrac { -1-\sqrt { 33 } }{ 4 }




(iii) 4x2+4√3x+3 = 0

Solution: The above equation can be written as

(2x)2+2. 2x.√3+(√3)2 = 0

(2x+√3)2 = 0

2x+√3 = 0 and 2x+√3 = 0

x = –\cfrac { \sqrt { 3 } }{ 2 } and x = –\cfrac { \sqrt { 3 } }{ 2 }

Hence the root are –\cfrac { \sqrt { 3 } }{ 2 }  and –\cfrac { \sqrt { 3 } }{ 2 }

 

(iv) 2x2+ x + 4 = 0

Solution: On Dividing both sides by 2

x2 + \cfrac { x }{ 2 } + 2 = 0

x2+2.x.\cfrac { 1 }{ 4 }  +{ \left( \cfrac { 1 }{ 4 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 4 } \right) }^{ 2 } +2 = 0

{ \left( x+\cfrac { 1 }{ 4 } \right) }^{ 2 }\cfrac { 1 }{ 16 } +2 = 0

{ \left( x+\cfrac { 1 }{ 4 } \right) }^{ 2 } + \cfrac { 31 }{ 16 } = 0

{ \left( x+\cfrac { 1 }{ 4 } \right) }^{ 2 } = –\cfrac { 31 }{ 16 } < 0

Now the value of { \left( x+\cfrac { 1 }{ 2 } \right) }^{ 2 } can’t be negative for any real value of x. So there is no real value of x satisfying the given equation. Hence the given equation has no real roots.

 

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Solution: Let ax2+bx+c = 0 be the quadratic equation.

If b2-4ac ≥ 0, then the roots of the quadratic equation is given by, x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } , this formula is called quadratic formula.

 

(i) 2x2 – 7x + 3 = 0

Solution: On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 2, b = -7, c = 3

Since b2-4ac = (-7)2-4.2.3 = 49- 24 = 25 > 0

Then the solution of the quadratic equation is given by,

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

\cfrac { -(-7)\pm \sqrt { 25 } }{ 2\times 2 }

= \cfrac { 7\pm 5 }{ 4 }

Taking the positive sign,

x = \cfrac { 7+5 }{ 4 } = \cfrac { 12 }{ 4 } = 3

And taking the negative sign,

x = \cfrac { 7-5 }{ 4 } = \cfrac { 2 }{ 4 } = \cfrac { 1 }{ 2 }

Hence the roots are 3, \cfrac { 1 }{ 2 } .

 

(ii) 2x2+ x – 4 = 0

Solution: On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 2, b = 1, c = -4

Since b2-4ac = (1)2-4.2.(-4) = 1+32 = 33 > 0

Then the solution of the quadratic equation is given by,

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

\cfrac { -1\pm \sqrt { 33 } }{ 2\times 2 }

\cfrac { -1\pm \sqrt { 33 } }{ 4 }

Taking the positive sign,

x = \cfrac { -1+\sqrt { 33 } }{ 4 }

And taking the negative sign,

x = \cfrac { -1-\sqrt { 33 } }{ 4 }

Hence the roots are \cfrac { -1+\sqrt { 33 } }{ 4 }  and \cfrac { -1-\sqrt { 33 } }{ 4 } .

 

(iii) 4x2+4√3x+3 = 0

Solution: On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 4, b = 4√3, c = 3

Since, b2-4ac = (4√3)2-4.4.3 = 48 – 48 = 0

Then the solution of the quadratic equation is given by,

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

= \cfrac { -4\sqrt { 3 } \pm 0 }{ 2\times 4 }

= –\cfrac { \sqrt { 3 } }{ 2 } and –\cfrac { \sqrt { 3 } }{ 2 }

Hence the root of the quadratic equation are –\cfrac { \sqrt { 3 } }{ 2 }  and –\cfrac { \sqrt { 3 } }{ 2 } .

 

(iv) 2x2+ x + 4 = 0

Solution: On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 2, b = 1, c = 4

Since, b2-4ac = (1)2– 4.2.4 = 1 – 32 = -31 ≱ 0

Hence the given quadratic equation has no real roots.

 

3. Find the roots of the following equations.

(i) x – \cfrac { 1 }{ x }  = 3, x ≠0

Solution: The given equation can be written as

x2 -1 = 3x

x2 -3x -1 = 0

On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 1, b = -3, c = -1

Since b2-4ac = (-3)2-4.1.(-1) = 9+4 = 13 > 0

Then the solution of the quadratic equation is given by,

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

= \cfrac { -(-3)\pm \sqrt { 13 } }{ 2\times 1 }

= \cfrac { 3\pm \sqrt { 13 } }{ 2 }

Hence the roots of the quadratic equation are \cfrac { 3+\sqrt { 13 } }{ 2 }  and \cfrac { 3-\sqrt { 13 } }{ 2 } .




(ii) \cfrac { 1 }{ x+4 }  – \cfrac { 1 }{ x-7 }  = \cfrac { 11 }{ 30 } , x ≠ -4, 7

Solution: The above equation can be written as

\cfrac { (x-7)-(x-4) }{ (x+4)(x-7) } = \cfrac { 11 }{ 30 }

= \cfrac { -11 }{ { x }^{ 2 }-3x-28 }  = \cfrac { 11 }{ 30 }

-30 = x2-3x-28

x2-3x+2 = 0

On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 1, b = -3, c = 2

Since b2-4ac = (-3)2-4.1.2 = 9-8 = 1 > 0

Then the solution of the quadratic equation is given by,

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

= \cfrac { -(-3)\pm \sqrt { 1 } }{ 2\times 1 }

= \cfrac { 3+1 }{ 2 } and \cfrac { 3-1 }{ 2 }

= 2 and 1

Hence the roots of the quadratic equation are 2 and 1

 

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is . Find his present age.

Solution: Let x be the present age of Rehman.

Then his age 3 years ago is x-3 and 5 years from now is x+5.

Hence by given condition,

\cfrac { 1 }{ x-3 } + \cfrac { 1 }{ x+5 } = \cfrac { 1 }{ 3 }

\cfrac { x+5+x-3 }{ (x-3)(x+5) } = \cfrac { 1 }{ 3 }

\cfrac { 2x+2 }{ { x }^{ 2 }+2x-15 } = \cfrac { 1 }{ 3 }

x2+2x-15 = 6x+6

x2-4x-21 = 0

On Comparing the above equation with the quadratic equation ax2+bx+c = 0 we have,

a = 1, b = -4, c = -21

Since b2-4ac = (-4)2-4.1.(-21) =16+84 = 100 > 0

Then the solution of the quadratic equation is given by,

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

= \cfrac { -(-4)\pm \sqrt { 100 } }{ 2\times 1 }

= \cfrac { 4\pm 10 }{ 2 }

Taking the positive value,

x = \cfrac { 4+10 }{ 2 } = 7

And taking the negative value,

x = \cfrac { 4-10 }{ 2 } = -3

Since, age cannot be negative, hence x ≠ -3

Hence the present age of Rehman is 7 year.

 

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.

Solution: Let the marks in mathematics be x. Then the marks in English is 30-x.

Using the given condition,

(x+2)(30-x-3)=210

(x+2) (27-x) =210

27x+54-x2-2x=210

x2-25x+156=0

x2-12x-13x+156=0

(x-12) (x-13) = 0

x-12=0

x=12

Or, x-13=0

x=13

If the marks in mathematics are 12, then marks in English will be 30 – 12 = 18

If the marks in mathematics are 13, then marks in English will be 30 – 13 = 17

6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution: Let the length of the shorter side is x meters. Then the length of longer side is x+30 meters

Hence, the length of the diagonal is \sqrt { { x }^{ 2 }+{ (x+30) }^{ 2 } }  meters

Therefore from the given condition,

\sqrt { { x }^{ 2 }+{ (x+30) }^{ 2 } }  = x+60

Squaring both side,

x2+(x+30)2 = (x+60)2

x2+x2+60x+900 =x2+120x+3600

x2-60x-2700 = 0

x2-90x+30x-2700 = 0

(x-90)(x+30)=0

x-90=0

x=90

Or, x+30=0

x = -30

Since the length cannot be negative, hence the length of shorter side is 90 meters and therefore the length of longer side is 90+30 =120 meters.

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.

Solution: Let the larger number is x then the smaller number is \sqrt { 8x }

Therefore using the given condition,

x2 – 8x = 180

x2-8x-180 = 0

x2-18x+10x-180 = 0

(x-18)(x+10) = 0

x-18 = 0

x = 18

Or, x+10 = 0

x = -10

If we take x = -10 then larger number is -10 and the smaller number is \sqrt { 8(-10) } , which is not possible.

Hence larger number is 18 and smaller number is \sqrt { 8\times 18 }  = ±12.

 

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.

Solution: Let the speed of the train is x km/h.

We know that,

time = \cfrac { distance }{ speed }

Then the time taken by the train to cover 360 km is = \cfrac { 360 }{ x } hours

When the speed is x+5, then times taken to cover 360 km is = \cfrac { 360 }{ x+5 } hours.

From the given conditions we have,

\cfrac { 360 }{ x }   – \cfrac { 360 }{ x+5 } = 1

360(x+5-x) = x(x+5)

x2+5x-1800 = 0

x2+45x-40x-1800 = 0

(x+45)(x-40) = 0

x+45 = 0

x = -45

Or, x-40 = 0

x = 40

Since speed can’t be negative then x = -45 is not possible. Therefore the speed of the train is 40 km/hour.




9. Two water taps together can fill a tank in 9\cfrac { 3 }{ 8 }  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.

Solution: Let the time taken by the smaller tap is x hour. Then in 9\cfrac { 3 }{ 8 } = \cfrac { 75 }{ 8 } hour the smaller tap fill the tank \cfrac { 75 }{ 8 } .\cfrac { 1 }{ x } = \cfrac { 75 }{ 8x } part.

And the larger tap takes x-10 hours to fill the tank separately. Then it will fill the tank in \cfrac { 75 }{ 8 }  hour is \cfrac { 75 }{ 8 } .\cfrac { 1 }{ x-10 }  = \cfrac { 75 }{ 8(x-10) }

Then by given condition,

\cfrac { 75 }{ 8x } + \cfrac { 75 }{ 8(x-10) }  = 1

\cfrac { 75 }{ 8 } (x-10+x) = x(x-10)

x2-10x = \cfrac { 75 }{ 8 } (2x-10)

x2-10x = \cfrac { 75 }{ 4 } (x-5)

4x2-40x = 75x-375

4x2-115x+375 = 0

Comparing the above quadratic equation by ax2+bx+c = 0 we get

a = 4, b = -115, c =375

Then the solution is given by

x = \cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

=\cfrac { -(-115)\pm \sqrt { { (-115) }^{ 2 }-4\times 4\times 375 } }{ 2\times 4 }

= \cfrac { 115\pm 85 }{ 8 }

x = \cfrac { 115+ 85 }{ 8 }  (taking the positive sign)

x = 25

Or, x = \cfrac { 115- 85 }{ 8 } (taking the negative sign)

x = \cfrac { 15 }{ 4 }

If time taken by smaller tap is \cfrac { 15 }{ 4 }  hours.

Then, the time taken by larger tap = \cfrac { 15 }{ 4 }  – 10 = –\cfrac { 25 }{ 4 } , which is not possible.

Therefore time taken by the smaller tap is 25 hours and the time taken by larger tap is 25-10 = 15 hours

 

10. An express train takes 1 hour less than a passenger train to travel 132 km between Maysore 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.

Solution: Let speed of the passenger train is x km/h then the speed of the express train is (x+11) km/h.

As we know that,

time = \cfrac { distance }{ speed }

Then the time taken by the passenger to travel 132 km is \cfrac { 132 }{ x }

and time taken by the express train to travel 132 km is \cfrac { 132 }{ x+11 }

Then by the given condition,

\cfrac { 132 }{ x } \cfrac { 132 }{ x+11 }  = 1

132(x+11-x) = x(x+11)

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 = 0

x = -44

Or, x-33 = 0

x = 33

Since the speed can’t be negative, then x = -44 not possible.

Therefore speed of the passenger train is 33 km/h and the speed of the express train is 33+11 = 44 km/h.

 

11. Sum of the areas of the two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of two squares.

Solution: Let side of 1st square is x m.

Then, perimeter = 4x m.

From the given condition the perimeter of the 2nd square is (4x+24) m

Then the side of 2nd square is \cfrac { 4x+24 }{ 4 } = (x+6) m

Area of 1st square is x2 m2

Area of 2nd square is (x+6)2 m2

From the given condition we have,

x2+(x+6)2 = 468

x2+ x2+12x+36 = 468

2x2+12x-432 = 0

x2+6x-216 = 0

x2+18x-12x-216 = 0

x(x+18)-12(x+18) = 0

(x+18)(x-12) = 0

x+18 = 0

x = -18

Or, x-12 = 0

x = 12

Since side can’t be negative then x = -18 is not possible.

Then the side of 1st square is 12 m and the 2nd square is (12+6) = 18 m.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 – Quadratic Equations, has been designed by the NCERT to test the knowledge of the student on the topic – Solution of a Quadratic Equation by Completing the Square

Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 – Quadratic Equations

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