Students can Download Class 10 Maths Chapter 10 Quadratic Equations Ex 10.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Ex 10.4

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

Answer:

(i) 2x^{2} – 3x + 5 = 0

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

b^{2} – 4ac = (-3)^{2} – 4(2) (5)

= 9 – 40 = -31

b^{2} – 4ac = -31 < 0

It has no real roots.

ii) 3x^{2} – 4\(\sqrt{3}\)x + 4 = 0

It is in the form of ax^{2} + bx + c = 0

a = 3, b = – 4\(\sqrt{3}\) and c = 4

Discriminant = b^{2} – 4ac

= (- 4\(\sqrt{3}\))^{2} – 4 × 3 × 4

= 16 × 3 – 16 × 3 = 48 – 48

Discriminant = 0

Hence, the given Quadratic equation has two equal real roots.

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

iii) 2x^{2} – 6x + 3 = 0

It is in the form of ax^{2} + bx + c = 0

a = 2, b = – 6 and c = 3

Discriminant = b^{2} – 4ac

= (- 6)^{2} – 4 × 2 × 3

= 36 – 24

Discriminant = 12

The given Quadratic equation has two distinct real roots.

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

Answer:

i) 2x^{2} + Kx + 3 = 0

It is in the form ax^{2} + bx +c = 0

a = 2, b = K and c = 3

b^{2} – 4ac = 0 [have equal root]

(K)^{2} – 4 (2) (3) = 0

K^{2} – 24 = 0

K^{2} = 24

K = ± \(\sqrt{6 \times 4}\)

K = ± 2\(\sqrt{6}\).

The required value of K are 2\(\sqrt{6}\) and – 2\(\sqrt{6}\)

ii) Kx (x – 2) + 6 = 0

Kx^{2} – 2Kx + 6 = 0

It is in the form ax^{2} + bx + c = 0

a = K, b = – 2K and c = 6

Nature of roots are equal

b^{2} – 4ac = 0

(- 2K)^{2} – 4 × K × 6 = 0

4K^{2} – 24K = 0

4K (K – 6) = 0

4K = 0 or K – 6 = 0

K = 0 and K = 6

The required values of K are K = 0 and 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.

Answer:

Let Breadth of mango grove be ‘x’m.

Breadth AD = BC = x m.

Area = 800 sq. m.

Length is twice its breadth.

∵ Length = 2x m.

length × breadth = Area of rectangle

2x × x = 800

2x^{2} = 800

x^{2} = 400

∴ x = ±20 m.

∴ It is possible to design a rectangular mango grove.

Breadth of mango grove, x = 20 m.

Length of mango, 2x = 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 the age of one friend is ‘x’ years and other friend is (20 – x)

Four years ago

The age of one friend is (x – 4) and other friend is (20 – x – 4) (or) (16 – x)

Product of their ages = 48

(x – 4) (16 – x) = 48

16x – 64 – x^{2} + 4x = 48

20x – x^{2} – 64 = 48

x^{2} – 20x + 48 + 64 = 0

x^{2} – 20x + 112 = 0

It is in the form ax^{2} + bx + c = 0

a = 1, b = – 20 and c = 112

It has no real roots.

∴ 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.

Answer:

Let the breadth of a rectangular park be ‘x’ m.

Its length is \(\frac{80-2 x}{2}\) m.

(∵ Perimeter = 80 m.)

Breadth × length = Area of rectangle.

\(x\left(\frac{80-2 x}{2}\right)\) = 400

(80 – 2x) = 800

80x – 2x^{2} = 800

-2x^{2} + 80x – 800 = 0

2x^{2} – 80x + 800 = 0

x^{2} – 40x + 400 = 0

(x – 20)^{2} =0

(x – 20) (x – 20) =0

∴ If x – 20 = 0, then x = 20

If x – 20 = 0, then x = 20

∴ Breadth of garden, x = 20 m.

Length = \(=\frac{80-2 x}{2}=\frac{80-40}{2}=\frac{40}{2}\) = 20 m

∴ Both length and breadth of garden are equal then it is not rectnagular but it is a square.

It is not possible to construct garden which has perimeter 80 m. and Area 400 m^{2}.