Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 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\(\sqrt {3}\) x + 4 = 0

(iii) 2x^{2} – 6x + 3 = 0

Solution:

(i) According to the question

2x^{2} – 3x + 5 = 0

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

a = 2, b = – 3, c = 5

∴ discriminant (D) = b^{2} – 4ac

= (-3)^{2} – 4 x 2 x 5

= 9 – 40

= – 31 < 0

Hence the given quadratic equation has no real roots.

(ii) According to the question

3x^{2} – 4\(\sqrt {3}\)x + 4 = 0

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

a = 3, b = -4\(\sqrt {3}\), c = 4

∴ discriminant (D) = b^{2} – 4ac

= (-4\(\sqrt {3}\))^{2} – 4 × 3 × 4

= 48 – 48 = 0

Hence the given quadratic equation has real and equal roots.

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

(iii) According to the question

2x^{2} – 6x + 3 = 0

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

a = 2, b = – 6, c = 3

∴ discriminant (D) = b^{2} – 4ac

= (-6)^{2} – 4 × 2 × 3

= 36 – 24

= 12 > 0

∴ The given quadratic equation has real and distinct roots.

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

Hence, the roots of the 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) According to the question 2x^{2} + kx + 3 = 0

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

a = 2, b = k, c = 3

∵ The roots of the given quadratic equation are equal

∴ discriminant (D) = 0

b^{2} – 4ac = 0

or (k)^{2} – 4 x 2 x 3 = 0

or k^{2} – 24 = 0

or k^{2} = 24

or k = +\(\sqrt {24}\)

or k = ± 2\(\sqrt {6}\)

Hence for roots to be equal we must have k = + 2\(\sqrt {6}\)

(ii) According to the question

kx (x – 2) + 6 = 0

or kx^{2} – 2kx + 6 = 0

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

a = k, b = – 2k, c = 6

∵ The roots of the given quadratic equation are equal

∴ D = 0

or b^{2} – 4ac = 0

or (- 2k)^{2} – 4 × k × 6 = 0

or 4k^{2} – 24k = 0

or 4k [k – 6] = 0

i.e., 4k = 0

or k – 6 = 0

⇒ k = 0 or k = 6

Putting k = 0 in the quadratic equation, we get 6 = 0 which is impossible.

Hence the correct value of k is only k = 6

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 the rectangular grove = x m

and the length of the rectangular grove = 2x m

Area of the rectangular grove

= Length × Breadth

= [2x × x] m^{2}

= 2x^{2} m^{2}

According to the question

2x^{2} = 800

⇒ x^{2} = \(\frac{800}{2}\) = 400

⇒ x = ±\(\sqrt {400}\)

⇒ x = ±20

∵ The length of the rectangular grove cannot be negative

Therefore leaving x = – 20

∴ x = 20

Here we have obtained a real value of x. This is the reason that it is possible to design such a rectangular mango grove.

∴ Breadth of the rectangular grove = 20 m

and length of the 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. Fouryears ago, the product of their ages in years was 48.

Solution:

Let the age of first friend = x years

Then, the age of second friend

= (20 – x) years

Four years ago,

Age of first friend = (x – 4) years

and Age of second friend = (20 – x – 4) years

= (16 – x) years

∴ Their product = (x – 4) (16 – x)

= 16x – x^{2} – 64 + 4x

= – x^{2} + 20x – 64

According to the 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)

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

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

∴ Discriminant (D) = b^{2} – 4ac

= (-20)^{2} – 4 × 1 × 112

= 400 – 448

= – 48 < 0

∴ The roots are not real.

Therefore, no real value of x can satisfy the quadratic equation (1)

Hence the 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 the length of the rectangular park = x m

and the breadth of the rectangular park

= y m

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

andarea of the rectangular park = xy m^{2}

According to the first condition,

2 (x + y) = 80

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

or y = 40 – x ….(1)

According to the second condition,

xy = 400

or x (40 – x) = 400 [Using (1)]

or 40x – x^{2} = 400

or 40x – x^{2} – 400 = 0

or x^{2} – 40x + 400 = 0

Comparing it with ax^{2} + bx + c – 0,

a = 1. b = – 40, c = 400

∴ discriminant (D) = b^{2} – 4ac

= (-40)^{2} – 4 × 1 × 400

= 1600 – 1600 = 0

Hence the roots of this quadratic equation will be real and equal.

When x = 20, then from (1)

y = 40 – 20 = 20

∴ Measure of length and breadth of the rectangular park is equal to 20 m.

Hence it is possible to design the rectangular park and it is a square.

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