Question

4 x square + 4 x minus bracket a square minus b square bracket close equal to zero

Answer 1

this is your answer......

Answer 2

Solution: $x=-\dfrac{a+b}{2},\dfrac{a-b}{2}$

NOTE:

The correct equation is

$4x^{2}+4bx-(a^{2}-b^{2})=0$

Step-by-step explanation:

Here, the given equation is

$4x^{2}+4bx-(a^{2}-b^{2})=0$

$\Rightarrow 4x^{2}+4bx-a^{2}+b^{2}=0$

• Hint. We have just opened the brackets using $(-)(+)=(-)$ and $(-)(-)=(+)$.

$\Rightarrow 4x^{2}+4bx+b^{2}-a^{2}=0$

• Hint. We have rearranged the terms so that we can use the algebraic identity $(a+b)^{2}=a^{2}+2ab+b^{2}$.

$\Rightarrow \{(2x)^{2}+2\times 2x\times b+(b)^{2}\}-a^{2}=0$

$\Rightarrow (2x+b)^{2}-(a)^{2}=0$

• Hint. Now we use the algebraic identity $a^{2}-b^{2}=(a+b)(a-b)$ to factorize the expression.

$\Rightarrow (2x+b+a)(2x+b-a)=0$

$\therefore$ either $2x+b+a=0$ or, $2x+b-a=0$

$\Rightarrow 2x=-b-a$ or, $2x=-b+a$

$\Rightarrow 2x=-(a+b)$ or, $2x=a-b$

$\Rightarrow x=-\dfrac{a+b}{2}$ or, $x=\dfrac{a-b}{2}$

Thus the required solution is

$\boxed{x=-\dfrac{a+b}{2},\dfrac{a-b}{2}}$

#SPJ3