AP 10th Class Maths Important Questions Chapter 5 Quadratic Equations

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 b² – 4ac ≥ 0, then write the roots of a quadratic equation ax² + bx + c = 0
Solution:
When b² – 4ac ≥ 0 then the roots of given quadratic equation ax² + 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 [x² – (α + β )x + αβ ] = [x² + \(\frac{5}{3}\)x – \(\frac{-2}{3}\)
the quadratic polynomial will be 3x² + 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 2x² – 4x + 3 = 0.
Solution:
Discriminant of standard quadratic equation
ax² + bx + c = 0 is b² – 4ac
Now comparing the given quadratic equation 2x² – 4x + 3 = 0 with stan¬dard form of quadratic equation.
We get a = 2, b = -4, c = 3 then its discriminant = b² – 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
⇒ x² – 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 x² – x – 30 and verify the rela¬tion between the zeroes and its co¬efficients.
Solution:
Given quadratic polynomial = x² – x – 30
⇒ x² – x – 30 = 0 = 0
x² – 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
9x² + 4 – 12x – 12x + 8 + 3 = 0
9x² – 24x +15 = 0
3x² – 8x + 5 = 0
3x² – 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
⇒ x² – 4x – 192 = 0
⇒ x² – 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 5x² – 7x – 6 = 0 by the method of completing the square.
Solution:
Given that 5x² – 7x – 6 = 0
5x² – 7x = 6
x² – \(\frac{7}{5}\)x = \(\frac{6}{5}\)
⇒ x².2.\(\frac{7}{10}\)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 3x² + 11x + 10 = 0 by method of – completing the Square.
Solution:
Given : 3x² + 11x + 10 = 0
Dividing both sides by 3.

Question 3.
Solve the Quadratic equation 9x² – 9x + 2 = 0 by the method of com-pleting the square.
Solution:
Given : 9x² – 9x + 2 = 0
⇒ x² – x + \(\frac{2}{9}\) = 0
⇒ x² – x = –\(\frac{2}{9}\)