Solve the following quadratic equation for x:
x2−2ax−(4b2−a2)=0
We have,x2−2ax−(4b2−a2)=0
⟹ \implies⟹ x2−2ax+a2−4b2=0
⟹ (x−a)2−(2b)2=0 [Using a2−b2a^2-b^2a2−b2 = (a+b)(a−b)(a+b)(a-b)(a+b)(a−b)]
∴(x−a+2b)(x−a−2b)=0
⇒ x=a−2b or a+2b
Hence, x=a−2bx=a-2bx=a−2b or a+2ba+2ba+2b