Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

## JAC Board 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:

1. 2x^{2} – 3x + 5 = 0

2. 3x^{2} – 4\(\sqrt{3}\)x + 4 = 0

3. 2x^{2} – 6x + 3 = 0

Solution:

The given equation is of the form

ax^{2 }+ bx + c = 0; where a = 2, b = -3 and c = 5.

Then, the discriminant

b^{2} – 4ac = (-3)^{2} – 4(2)(5)

= 9 – 40

= -31 <0

So, the given equation has no real roots.

2. Comparing the given equation with the standard quadratic equation

ax^{2} + bx + c = 0, we get a = 3, b = -4\(\sqrt{3}\) and c = 4.

Then, the discriminant

b^{2} – 4ac = (-4\(\sqrt{3}\))^{2} – 4(3)(4)

= 48 – 48

= 0

So, the given equation has equal real roots.

The roots are \(-\frac{b}{2 a},-\frac{b}{2 a}\)

i.e., \(-\frac{-4 \sqrt{3}}{2(3)},-\frac{-4 \sqrt{3}}{2(3)}\), i.e., \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. The given equation is of the form

ax^{2} + bx + c = 0, where a = 2, b = -6 and c = 3.

Then, the discriminant

b^{2} – 4ac = (-6)^{2} – 4(2)(3)

= 36 – 24

= 12

So, the given equation has two distinct roots.

The roots are given by

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

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

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

Thus, the roots of the given 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:

1. 2x^{2} + kx + 3 = 0

2. kx (x – 2) + 6 = 0

Solution:

1. Comparing the given equation with the standard quadratic equation, we have

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

Then, the discriminant = b^{2} – 4ac

= (k)^{2} – 4 (2) (3)

= k^{2} – 24

If the equation has two equal roots, then the discriminant = 0

∴ k^{2} – 24 = 0

∴ k^{2} = 24

∴ k = ±\(\sqrt{24}\)

∴ k ± 2\(\sqrt{6}\)

2. kx(x – 2) + 6 = 0

∴ kx^{2} – 2kx + 6 = 0

Here, a = k, b = -2k and c = 6.

Then, the discriminant = b^{2} – 4ac

= (-2k)^{2} – 4(k)(6)

= 4k^{2} – 24k

If the equation has two equal roots, then the discriminant = 0

∴ 4k^{2} – 24k = 0

∴ 4k (k – 6) = 0

∴ k = 0 or k = 6

But k = 0 is not possible because if k = 0, the equation reduces to 6 = 0, not a quadratic equation.

∴ 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:

Considering that the required mango grove can be designed, let the breadth of the mango grove be x m.

Then, the length of the mango grove is 2x m. Area of rectangular mango grove

= Length × Breadth

= 2x × x

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

The required area is 800 m^{2}.

∴ 2x^{2} = 800

∴ 2x^{2} – 800 = 0

∴ x^{2} – 400 = 0

If the above equation has real roots, then it is possible to design the required mango grove.

Here, a = 1, b = 0 and c = -400.

Then, the discriminant = b^{2} – 4ac

= (0)^{2} – 4(1)(-400)

= 1600 > 0

Hence, the equation has real roots. So, it is possible to design the mango grove with required measures.

Now, x^{2} – 400 = 0

∴ (x + 20) (x – 20) = 0

∴ x + 20 = 0 or x – 20 = 0

∴ x = -20 or x = 20

Since x is the breadth of the rectangular mango grove, x = -20 is not possible.

∴ x = 20 and 2x = 40

Thus, the length of the mango grove is 40 m and its breadth is 20 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 the present ages of two friends be x years and (20 – x) years.

Four years ago, their respective ages were (x – 4) years and (20 – x – 4) years, i.e., (16 – x) years.

Then, according to given,

(x – 4) (16 – x) = 48

∴ 16x – x^{2} – 64 + 4x = 48

∴ -x^{2} + 20x – 64 – 48 = 0

∴ -x^{2} + 20x – 112 = 0

∴ x^{2} – 20x + 112 = 0

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

Then, the discriminant

b^{2} – 4ac = (-20)^{2} – 4(1)(112)

= 400 – 448

= -48 < 0

Hence, the equation has no real roots. So, the given situation is not possible.

Question 5.

Is it possible to design a rectangular park of perimeter 80 m and area 400 m^{2} ? If so, find its length and breadth.

Solution:

Let the length of the rectangular park be x m.

Perimeter of rectangular park = 2 (Length + Breadth)

∴ 80 = 2(x + Breadth)

∴ 40 = x + Breadth

∴ Breadth = (40 – x) m

Now, area of rectangular park = Length × Breadth

∴ 400 = x (40 – x)

∴ 400 = 40x – x^{2}

∴ x^{2} – 40x + 400 = 0

Here, a = 1, b = -40 and c = 400.

Then, the discriminant = b^{2} – 4ac

= (-40)^{2} – 4(1)(400)

= 1600 – 1600

= 0

Hence, the equation has equal real roots. So, it is possible to design a rectangular park with given measures.

x^{2} – 40x + 400 = 0

∴ x^{2} – 20x – 20x + 400 = 0

∴ x(x – 20) – 20(x – 20) = 0

∴ (x – 20) (x – 20) = 0

∴ x – 20 = 0 or x – 20 = 0

∴ x = 20 or x = 20

Thus, the length of the rectangular park = x = 20 m and the breadth of the rectangular park = 40 – x = 40 – 20 = 20 m.

Note: Here the shape of the park turns out to be a square, but as we know, every square is a rectangle.