solve the following by quadratic formula 4x2-4ax+(a2-b2)
Answers
Answered by
80
4x² +4bx -(a² -b²) = 0
4x² +4bx -a² + b² =0
(2x)² + 2.(2x).b + b² -a² =0
(2x + b)² -a² = 0
use formula,
a²-b² = (a -b)(a + b)
{2x + b -a}{2x + b +a} =0
x = (a - b)/2 , - (a + b)/2
4x² +4bx -a² + b² =0
(2x)² + 2.(2x).b + b² -a² =0
(2x + b)² -a² = 0
use formula,
a²-b² = (a -b)(a + b)
{2x + b -a}{2x + b +a} =0
x = (a - b)/2 , - (a + b)/2
Answered by
61
Answer:
Step-by-step explanation:
Solution :-
Here, we have
4x² - 4ax + (a² - b²) = 0
Constant term = (a² - b²) = (a - b) (a + b)
Coefficient of middle term = - 4a
Coefficient of middle term = - 4a = - [2(a + b) + 2(a - b)]
Therefore,
⇒ 4x² - 4ax + (a² - b²) = 0
⇒ 4x² - [2(a + b)x - 2(a - b)]x + (a + b) (a - b) = 0
⇒ 4x² - 2(a + b)x - 2(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)] [2x - (a - b)] = 0
⇒ [2x - (a + b)] = 0 or [2x - (a - b)] = 0
⇒ 2x = a + b or 2x = a - b
⇒ x = a + b/2, a - b/2
Hence, x = a + b/2, a - b/2
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