## NCERT solutions for class 10 Maths

**Chapter 4**

## Quadratic Equations

**Exercise 4.4**

**Ex 4.4 Question 1.**

**Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2 x^{2} – 3x + 5 = 0
(ii) 3x^{2} – 4√3x + 4 = 0
(iii) 2x^{2} – 6x + 3 = 0**

**Solution:**

**[Comparing the equation with ax^{2} + bx + c = 0,]**

(i) *2x*^{2} – 3*x* + 5 = 0

*a* = 2, *b* = -3 and *c* = 5

Discriminant (D) = *b*^{2} – 4*ac*

*= *( – 3)^{2} – 4 (2) (5) = 9 – 40

= – 31

As you can see, b^{2} – 4ac < 0

Therefore, the roots are imaginary.

(ii) 3*x*^{2} – 4√3*x* + 4 = 0

*a* = 3, *b* = -4√3 and *c* = 4

Discriminant (D) = *b*^{2} – 4*ac*

= (-4√3)^{2 }– 4(3)(4)

= 48 – 48 = 0

As *b*^{2} – 4*ac* = 0,

Hence, the roots are real and equal.

Using formula

x* *= -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots of the equation are 2/√3 and 2/√3.

(iii) 2*x*^{2} – 6*x* + 3 = 0

*a* = 2, *b* = -6, *c* = 3

Discriminant (D) = *b*^{2} – 4*ac*

= (-6)^{2} – 4 (2) (3)

= 36 – 24 = 12

As *b*^{2} – 4*ac* > 0,

Therefore, the roots are distinct and real.

= (-(-6) ± √(-6^{2}-4(2)(3)) )/ 2(2)

= (6±2√3 )/4

= (3±√3)/2

Hence, the roots for the equation are (3+√3)/2 and (3-√3)/2

**Ex 4.4 Question 2.**

** Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x^{2} + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0**

**Solutions:**

(i) 2*x*^{2} + *kx* + 3 = 0

*a* = 2, *b* = k and *c* = 3

Discriminant (D) = *b*^{2} – 4*ac*

= (*k*)^{2} – 4(2) (3)

= *k*^{2} – 24

For equal roots,

Discriminant = 0

*k*^{2} – 24 = 0

*k*^{2} = 24

k = ±√24 = ±2√6

(ii) *kx*(*x* – 2) + 6 = 0

or *kx*^{2} – 2*kx* + 6 = 0

*a* = *k*, *b* = – 2*k* and *c* = 6

Discriminant (D) = *b*^{2} – 4*ac*

= ( – 2*k*)^{2} – 4 (*k*) (6)

= 4*k ^{2}* – 24

*k*

For equal roots,

*b*^{2} – 4*ac* = 0

4*k*^{2} – 24*k* = 0

4*k* (*k* – 6) = 0

Either 4*k* = 0 or *k* = 6 = 0

*k* = 0 (not possible)

*k* = 6

**Ex 4.4 Question 3.**

** Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m ^{2}? If so, find its length and breadth.**

**Solution:**

Let the breadth of mango grove = x m

∴The length = 2x.

Area = (2x) (x)= 2x^{2}

2x^{2 }= 800

*x*^{2 }= 800/2 = 400

*x*^{2 }– 400 =0

*a* = 1, *b* = 0, *c* = 400

As we know, Discriminant = *b*^{2} – 4*ac*

=> (0)^{2} – 4 × (1) × ( – 400) = 1600

Here, *b*^{2} – 4*ac* > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

*l *= ±20 [the length cannot be negative.]

Therefore, breadth = 20 m

Length = 2 × 20 = 40 m

**Ex 4.4 Question 4.**

**Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.**

**Solution:**

Let the age of first friend = x years.

Then, the age of the other friend = (20 – x) years.

Four years ago,

Age of First friend = (*x* – 4) years

Age of Second friend = (20 – *x* – 4) = (16 –* x*) years

(*x* – 4) (16 – *x*) = 48

16*x – x*^{2} – 64 + 4*x* = 48

* – x*^{2} + 20*x – *112 = 0

*x*^{2} – 20*x + *112 = 0

*a* = *1*, *b* = -2*0* and *c* = 112

Discriminant (D) = *b*^{2} – 4*ac*

*= (-*20*)*^{2} – 4 × 112

= 400 – 448 = -48

*b*^{2} – 4*ac *< 0

Therefore, there is no real roots of this quadratic equation. Hence, this situation is not possible.

**Ex 4.4 Question 5.**

** Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.**

**Solution:**

Let the length of the park= x

Perimeter of the rectangular park = 80 m

The breadth of the park= 40 – x m [Perimeter= 2 (*l + b*) ]

Perimeter of the rectangular park = 2 (*l + b*) = 80

40x* *– x^{2 }= 400

*x*^{2}* *–^{ }40x + 400* *= 0,

*x*^{2}* – *20*x – *20*x + *400 = 0

*x *(*x *– 20) – 20 (*x *– 20) = 0

(*x *– 20) (*x* – 20) = 0

(*x *– 12)² = 0 or (*x* – 13) = 0

(*x *– 20) = 20

⇒ x = 20

Hence, the length = 20 m

And breadth of the park = 40 – 20 = 20 m.