Question

Solve for x : x^2 + 6x - (a^2 + 2a - 8 ) = 0

Answer 1

$\bold{x=(a-2) \text { or }(-4-a)}$ is the value of $\bold{x^{2}+6 x-\left(a^{2}+2 a-8\right)=0.}$

Given:

$x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$

To find:

Value of x=?

Solution:

To find the value of “a”, we first find the roots of the equation, for that  

$x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$

First solve the equation $a^{2}+2 a-8$

Using separation method, we get the value of the equation

$a^{2}+2 a-8=(a-2)(a+4)$

Putting the value of the equation$a^{2}+2 a-8 as (a-2)(a+4)$ in

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore putting the values we get  

$x=\frac{-6 \pm \sqrt{6^{2}-4.1 \cdot(a-2)(a+4)}}{2}$

After Solving the equation by putting the value of c, we get the values or roots of x as

$x=(a-2) \text { or }(-4-a)$

Therefore, the answer to the equation is that the value of “a” can be $\bold{x=(a-2) \text { or }(-4-a).}$

Answer 2

Answer:

x = -(a+4) Or x = (a-2)

Step-by-step explanation:

Given quadratic equation:

x²+6x-(a²+2a-8)=0

=> x²+6x-(a²+4a-2a-8)=0

=> x²+6x-[a(a+4)-2(a+4)]=0

=> x²+6x-(a+4)(a-2)=0

Splitting the middle term,we get

=>x²+(a+4)x-(a-2)x-(a+4)(a-2)=0

=> x[x+(a+4)]-(a-2)[x+(a+4)]=0

=> [x+(a+4)][x-(a-2)]=0

=> x+(a+4)=0 Or x-(a-2)=0

=> x = -(a+4) Or x = (a-2)

Therefore,

x = -(a+4) Or x = (a-2)

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