x^2-2ax-(4b^2-a^2)=0 solve for x using quadratic equation
Answers
bro here is your answer
x2 - 2ax - (4b2 - a2) = 0
Using the splitting middle term - the middle term of the general equation
is divided in two such values that:
Product = a.c
For the given equation a = 1 b = - 2a c = - (4b2 - a2)
= 1. - (4b2 - a2)
= - (4b2 - a2)
And either of their sum or difference = b
= - 2a
Thus the two terms are (2b - a) and - (2b + a)
Difference = 2b - a - 2b - a
= - 2a
Product = (2b - a) - (2b + a)
(∵ using a2 - b2 = (a + b) (a - b))
= - (4b2 - a2)
x2 - 2ax - (4b2 - a2) = 0
⇒ x2 + (2b - a)x - (2b + a)x - (2b - a)(2b + a) = 0
⇒ x[x + (2b - a)] - (2b + a)[x + (2b - a)] = 0
⇒ [x + (2b - a)] [x - (2b + a)] = 0
⇒ [x + (2b - a)] = 0 or [x - (2b + a)] = 0
⇒ x = (a - 2b) or x = (a + 2b)
Hence the roots of given equation are (a - 2b) or x = (a + 2b)
Given:
x²-2ax-(4b²- a²)= 0
To Find
the value of x
Solution:
⇒ x= -b± √b²- 4ac/2a
⇒ x= -(-2a) ± √(-2a)²- 4(b² + a²)/2.1
⇒ x= 2a ± √4a² + 16b² + 4a²/2
⇒ x= 2a ± √16b²/2
⇒ x= 2a ± 4b/2
⇒ x= 2(a ± 2b)/2
⇒ x= a ± 2b
Therefore, the value of x is a ± 2b.