Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

Question 1.

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i)2x^{2} – 3x + 5 = 0

(ii) 3x^{2} – 4√3x + 4 = o

(iii) 2x^{2} – 6x + 3 = 0

Solution:

(i) Given quadratic equation is, 2x^{2} – 3x + 5 = 0

Compare it with ax^{2} + bx + c = 0

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

D = b^{2} – 4ac

= (-3)^{2} 4 × 2 × 5

= 9 – 40 = -31 < 0

Hence, given quadratic equation has no real roots.

(ii) Given quadratic equation is, 3x^{2} – 4√3x + 4 = 0

Compare it with ax^{2} + bx + c = 0

a = 3, b = -4√3, c = 4

D = b^{2} – 4ac

= (-4√3)^{2} – 4 × 3 × 4

= 48 – 48 = 0

given equation has real and equal roots.

Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-4 \sqrt{3}) \pm \sqrt{0}}{2 \times 3}\) = \(\frac{2}{\sqrt{3}}\)

Hence, roots of given quadratic equation are \(\frac{2}{\sqrt{3}}\) and \(\frac{2}{\sqrt{3}}\).

(iii) Given quadratic equation is :

2x^{2} – 6x + 3 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 2, b = -6, c = 3

D = b^{2} – 4ac

= (-6)^{2} 4 × 2 × 3

= 36 – 24 = 12 > 0

∴ given equation has real and distinct roots.

Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-6) \pm \sqrt{12}}{2 \times 2}\)

= \(\frac{6 \pm 2 \sqrt{3}}{4}\)

= \(\frac{3 \pm \sqrt{3}}{2}\)

= \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\)

Hence, roots of given quadratic equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3+\sqrt{3}}{2}\).

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

Solution:

(i) Given quadratic equation is : 2x^{2} + kx + 3 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 2, b = k, c = 3

∵ roots of the given quadratic equation are equal.

∴ D = 0

b^{2} – 4ac = 0

Or(k)^{2} – 4 × 2 × 3 = 0

Or k^{2} – 24 = 0

Or k^{2} = 24

Or k = ±√24

Or k = ±2√6.

(ii) Given quadratic equation is:

kx (x – 2) + 6 = 0

Or k – 2kx + 6 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = k, b = -2k, c = 6

∵ roots of the given quadratic equation are equal

∴ b^{2} – 4ac = 0

Or(-2k)^{2} – 4 × k × 6 = 0

Or 4k^{2} – 24k = 0

Or 4k[k – 6]= 0

Either 4k = 0 Or k- 6 = 0

k = 0 Or k = 6

∴ k = 0, 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.

Solution:

Let breadth of rectangular grove = x m

and length of rectangular grove = 2x m

Area of rectangular grove = length × breadth

= [x × 2x] m^{2} = 2 × 2 m^{2}

According to question

2x^{2} = 800

x^{2} = \(\frac{800}{2}\) = 400

x = ± √400

x = ± 20.

∵ length of rectangle cannot be negative.

So, we reject x = -20

∴ x = 20

∴ breadth of rectangular grove = 20 m

and length of rectangular grove = (2 × 20) m = 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.

Solution:

Let age of one friend = x years

and age of 2nd friend = (20 – x) years

Four years ago,

Age of 1st friend = (x – 4) years

Age of 2nd friend = (20 – x – 4) years = (16 – x) years

∴ Their product = (x – 4) (16 – x)

= 16x – x2 – 64 + 4x

= – x^{2} + 20x – 64

According to Question

– x^{2} + 20x – 64 = 48

Or – x^{2} + 20x – 64 – 48 = 0

Or – x^{2} + 20x – 112 = 0

Or x^{2} – 20x + 112 = 0 …………….(1)

Compare it with ax^{2} + bx + c = 0

∴ a = 1, b = -20, c = 112

D = b^{2} – 4ac

= (-20)^{2} – 4× 1 × 112

= 400 – 448 = -48 < 0

∴ roots are not real

then no real value of x satisfies the quadratic equation (1).

Hence, given situation is not possible.

Question 5.

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

Solution:

Let length of rectangular park = x m

Breadth of rectangular park = y m

∴ Perimeter of rectangular park = 2 (x + y) m

and area of rectangular park = xy m^{2}

According to 1st condition

2 (x + y) = 80

x + y = \(\frac{80}{2}\) = 40

y = 40 – x …………(1)

According to 2nd condition,

xy = 400

x (40 – x) = 400 [using (1)]

Or 40x – x^{2} = 400

Or 40x – x^{2} – 400 = 0

Or x^{2} – 40x + 400 = 0

Compare it with ax^{2} +bx + c = 0

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

D = b^{2} – 4ac

= (-40)^{2} – 4 × 1 × 400

= 1600 – 1600 = 0

Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-40) \pm \sqrt{0}}{2 \times 1}\)

= \(\frac{40}{2}\) = 20

When x = 20 then from (1)

y = 40 – 20 = 20

∴ Length and breadth of rectangular park are equal of measure 20 m.

Hence, given rectangular park exist and it is a square.