NCERT Solutions class 10 maths chapter 4

Quadratic Equations

NCERT Solutions class 10 maths chapter 4 Exercise 4.4

Question 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 – 4√3x + 4 = 0
(iii) 2x² – 6x + 3 = 0

Answer (i)

2x² – 3x + 5 = 0

On Comparing equation with ax² + bx + c = 0,

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

Discriminant = b² – 4ac

= ( – 3)² – 4 (2) (5) = 9 – 40

= – 31< 0

As b² – 4ac < 0

Hece no real root is possible for 2x² – 3x + 5 = 0.

Answer (ii)

3x² – 4√3x + 4 = 0

On Comparing equation with ax² + bx + c = 0

a = 3, b = -4√3 and c = 4

Discriminant = b² – 4ac

= (-4√3)² – 4(3)(4)

= 48 – 48 = 0

As b² – 4ac = 0,

given equation has two equal root and real root .

So, the roots will be –b/2a and –b/2a.

–b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Hence the roots are 2/√3 and 2/√3.

Answer (iii)

2x² – 6x + 3 = 0

On Comparing equation with ax² + bx + c = 0

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

Discriminant = b² – 4ac

= (-6)² – 4 (2) (3)

= 36 – 24 = 12

As b² – 4ac > 0,

Hence, the equation, has two distinct real roots 2x² – 6x + 3 = 0.

=

= (6±2√3 )/4

= (3±√3)/2

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

Question 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

Answer

(i) 2x² + kx + 3 = 0

On Comparing equation with ax² + bx + c = 0

a = 2, b = k and c = 3

Discriminant = b² – 4ac

= (k)² – 4(2) (3)

= k² – 24

For equal roots,

Discriminant = 0

k² – 24 = 0

k² = 24

k = ±√24 = ±2√6

Answer (ii)

kx(x – 2) + 6 = 0

or kx² – 2kx + 6 = 0

On Comparing equation with ax² + bx + c = 0

a = k, b = – 2k and c = 6

Discriminant = b² – 4ac

= ( – 2k)² – 4 (k) (6)

= 4k² – 24k

For equal roots

b² – 4ac = 0

4k² – 24k = 0

4k (k – 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

Question 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.

Answer

Let the breadth of mango grove be x.

Length of mango grove will be 2x.

Area of mango grove = (2x) (x)= 2x²

2x² = 800

x² = 800/2 = 400

x² – 400 =0

x = ±20

the value of length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

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.

Answer

Let’s the age of one friend be x years.

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

Four years ago,

Age of First friend = (x – 4) years

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

According to question

(x – 4) (16 – x) = 48

16x – x² – 64 + 4x = 48

– x² + 20x – 112 = 0

x² – 20x + 112 = 0

On Comparing the equation with ax² + bx + c = 0

a = 1, b = -20 and c = 112

Discriminant = b² – 4ac

= (-20)² – 4 × 112

= 400 – 448 = -48

b² – 4ac < 0

Therefore, there will be no real root possible for the equations. Hence, condition doesn’t exist.

Question 5

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

Answer

Let the length and breadth of the park be l and b.

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

So, l + b = 40

Or, b = 40 – l

Area of the rectangular park = l×b = l(40 – l) = 40l – l² = 400

l² – 40l + 400 = 0, A quadratic equation.

ON Comparing the equation with ax² + bx + c = 0

a = 1, b = -40, c = 400

Discriminant = b² – 4ac

=(-40)² – 4 × 400

= 1600 – 1600 = 0

Thus, b² – 4ac = 0

Therefore, this equation has equal real roots. So the situation is possible.

l = –b/2a

l = (40)/2(1) = 40/2 = 20

Therefore, length of rectangular park, l = 20 m

And breadth of the park, b = 40 – l = 40 – 20 = 20 m.