These AP 10th Class Maths Chapter Wise Important Questions Chapter 5 Quadratic Equations will help students prepare well for the exams.

## AP State Syllabus 10th Class Maths 5th Lesson Important Questions and Answers Quadratic Equations

Question 1.

If b^{2} – 4ac ≥ 0, then write the roots of a quadratic equation ax^{2} + bx + c = 0

Solution:

When b^{2} – 4ac ≥ 0 then the roots of given quadratic equation ax^{2} + bx + c = 0 are

\(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and \(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\)

Question 2.

Find the Quadratic polynomial with zeroes -2 and \(\frac{1}{3}\)

Solution:

Let α = -2; β = \(\frac{1}{3}\)

= -2 + \(\frac{1}{3}\) = \(\frac{-6+1}{3}=\frac{-5}{3}\)

α.β = -2. \(\frac{1}{3}\) = \(\frac{-2}{3}\)

Quadratic polynomial is [x^{2} – (α + β )x + αβ ] = [x^{2} + \(\frac{5}{3}\)x – \(\frac{-2}{3}\)

the quadratic polynomial will be 3x^{2} + 5x – 2

Question 3.

Two angles are complementary and one angle is 18° more than tne other, then find angles.

Solution:

Let smaller angle be x°

bigger angle be y°

Since these two angles are comple-mentary

x + y = 90° ………… (1)

Since bigger angle is more than smaller angle

by 18°, y – x = 18° ………….(2)

By solving (1) & (2), we get x = 36°, y = 54°

Question 4.

Find the discriminant of the quadratic equation 2x^{2} – 4x + 3 = 0.

Solution:

Discriminant of standard quadratic equation

ax^{2} + bx + c = 0 is b^{2} – 4ac

Now comparing the given quadratic equation 2x^{2} – 4x + 3 = 0 with stan¬dard form of quadratic equation.

We get a = 2, b = -4, c = 3 then its discriminant = b^{2} – 4ac

= (-4)^{2} – 4(2) (3)

= 16 – 24 = -8

∴ Discriminant = -8.

Question 1.

Find the roots of x + \(\frac { 6 }{ x }\) = 7, x ≠ 0

Solution:

x + \(\frac { 6 }{ x }\) = 7 ⇒ \(\frac{x^{2}+6}{x}\) = 7

⇒ x^{2} – 7x + 6 =0

⇒ (x – 6) (x – 1) = 0

⇒ x = 6 or 1

Roots = 6, 1

Question 2.

Length of a rectangle is 2 units greater than its breadth. If the area of the rect-angle is 120 sq. units then find its length.

Solution:

Let breadth of the rectangle = x

Length = x + 2 ,

Area = 120 sq. units

x(x+2) = 120

x2 + 2x- 120 = 0

(x + 12) (x- 10) = 0

x = – 12 or 10

Breadth cannot be negative

∴ Breadth of the rectangle = x

= 10 units

∴ Length = x + 2 = 10 + 2 = 12 units

Question 3.

Find the zeroes of the quadratic poly-nomial x^{2} – x – 30 and verify the rela¬tion between the zeroes and its co¬efficients.

Solution:

Given quadratic polynomial = x^{2} – x – 30

⇒ x^{2} – x – 30 = 0 = 0

x^{2} – 6x + 5x – 30 = 0

⇒ x(x – 6) + 5(x – 6) = 0

⇒ (x – 6) (x + 5) = 0

⇒ x – 6 = 0

x = 6

x + 5 = 0 x = -5

∴ Zeroes are α = 6 and β = – 5

Sum of zeroes = α + β = \(\frac{-\mathrm{b}}{\mathrm{a}}\)

⇒ 6 – 5 = \(\frac{-(-1)}{1}\)

⇒ 1 = 1

Product of zeroes α + β = 6(-5) = \(\frac{\mathrm{c}}{\mathrm{a}}\)

= -30 = \(\frac{-30}{1}\)

Hence the relation was verified.

Question 4.

Find the roots of the quadratic equa¬tion (3x – 2)^{2} – 4(3x – 2) + 3 = 0.

Solution:

(3x – 2)^{2} – 4(3x – 2) + 3 = 0

9x^{2} + 4 – 12x – 12x + 8 + 3 = 0

9x^{2} – 24x +15 = 0

3x^{2} – 8x + 5 = 0

3x^{2} – 3x – 5x + 5 = 0

3x(x – 1) – 5 (x – 1) = 0

(x- 1) (3x – 5) = 0

x = 1 (or) x = \(\frac{5}{3}\)

Roots of quadratic equation are 1, \(\frac{5}{3}\).

Question 5.

Two numbers differ by 4 and their product is 192. Find the numbers.

Solution:

Let the larger number be ‘x’

Since the difference of two numbers is 4,

Then the smaller number is (x – 4)

Their product = x(x – 4)

Given that product = 192

∴ x(x – 4) = 192

⇒ x^{2} – 4x – 192 = 0

⇒ x^{2} – 16x + 12x- 192 = 0

⇒ x(x – 16) + 12(x – 16) = 0

⇒ (x – 16)(x + 12) = 0

⇒ x = 16 or x = -12

If x = 16, then x – 4 = 16 – 4 = 12

then the numbers are 16 and 12.

If x = – 12, thenx-4 = – 12 – 4 = – 16

then the numbers are – 12 and – 16.

Question 1.

Find the roots of the equation 5x^{2} – 7x – 6 = 0 by the method of completing the square.

Solution:

Given that 5x^{2} – 7x – 6 = 0

5x^{2} – 7x = 6

x^{2} – \(\frac{7}{5}\)x = \(\frac{6}{5}\)

⇒ x^{2}.2.\(\frac{7}{10}\)x = \(\frac{6}{5}\)

adding \(\frac{49}{100}\) on the both sides

The roots of the quadratic equation are 2 and \(\frac{-3}{5}\)

Question 2.

Find the roots of the quadratic equation 3x^{2} + 11x + 10 = 0 by method of – completing the Square.

Solution:

Given : 3x^{2} + 11x + 10 = 0

Dividing both sides by 3.

Question 3.

Solve the Quadratic equation 9x^{2} – 9x + 2 = 0 by the method of com-pleting the square.

Solution:

Given : 9x^{2} – 9x + 2 = 0

⇒ x^{2} – x + \(\frac{2}{9}\) = 0

⇒ x^{2} – x = –\(\frac{2}{9}\)