Page 3 - CHAPTER 4 (Quadratic equations)

CHAPTER 4
QUADRATIC EQUATIONS

Solution: i. The given equation is (x + 1)² = 2(x – 3)

⟹ x² + 2x + 1 = 2x – 6 ⟹ x² + 2x – 2x + 1 + 6 = 0

⟹ x² + 7 = 0 ⟹ x² + 0. x + 7 = 0, which is of the form ax² + bx + c = 0

Hence, (x + 1)² = 2(x – 3) is a quadratic equation.

ii. Equation given is (x – 2) (x + 1) = (x – 1) (x + 3)

⟹ x² + x – 2x – 2 = x² + 3x – x – 3 ⟹ x² - x² - x – 2x – 2 + 3 = 0 ⟹ -3x + 1 = 0,

Which is not of the general form ax² + bx + c = 0

So, (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.

iii. Equation given is (x – 3) (2x + 1) = x (x + 5)

⟹ 2x² + x – 6x – 3 = x² + 5x ⟹ 2x² - x² - 5x – 5x -3 = 0 ⟹ x² - 10x – 3 = 0,

Which is of the general form ax² + bx + c = 0.

So, (x – 3) (2x + 1) = x(x + 5) Is a quadratic equation.

Roots of a quadratic equation:

Let f(x) = 0 is a quadratic equation, then x = a is called root or zero of f(x)

= 0. if it satisfies the equation.

i.e., if f(a) = 0 then x = a is called root or zero of f(x) = 0.

Example 2: In each of the following question, determine whether the

given values are the solution of the given equation or not:

2
1
5
i. - +2=0;x=5,x=
x² x 2
b
a
ii. a²x² - 3abx + 2b² = 0; x= ,x=
b a
1
Solution: i. After Putting = 5 and = in the equation.
2
2 5 2 5
- +2 and
(5)² 5 2 - 1 +2
( ) ² ( ) ²
2
2

3