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Quadratic Equations

Question
CBSEENMA10009620
Wired Faculty App

Solve the following quadratic equation for x,

4x2 + 4bx - (a2 - b2) = 0

Solution

Given equation,
4x2 + 4bx - (a2 - b2) = 0 
⇒ 4x2  +4bx - ( a + b) (a - b) = 0
⇒ 4x2 + 2[(a + b) - (a - b)] x - (a + b)(a-b) = 0
⇒ 4x2 +2 (a + b)x - 2(a - b)x - (a + b ) (a - b) = 0
⇒ 2x [2x + (a + b)] [(a - b)[2x + (a + b)] = 0
⇒ 2x + (a + b) = 0 or 2x - (a - b) = 0
⇒ x =  - (a + b)/2 OR x = (a - b)/2

Some More Questions From Quadratic Equations Chapter

Check whether the following are quadratic equations:
(x + 1)2 = 2(x – 3)

Check whether the following are quadratic equations:
x2 – 2x = (–2) (3 – x)

Check whether the following are quadratic equations:
(x – 2) (x + 1) = (x – 1) (x + 3)

Check whether the following are quadratic equations:
(x – 3) (2x + 1) = x(x + 5)

Check whether the following are quadratic equations:
(2x –1) (x–3) = (x + 5)(x–1)

Check whether the following are quadratic equations:
x2 + 3x + 1 = (x – 2)2

Check whether the following are quadratic equations:
(x + 2)3 = 2x(x2 – 1)

Check whether the following are quadratic equations:
x3 – 4x2 – x + 1 = (x – 2)3.

Represent the following problem situations in the form of quadratic equations:
The product of two consecutive positive integers is 306. We need to find the integers.

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