Question

Solve for x :

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

Answer 1

Required Answer:-

Given:

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

To find:

• The value of x.

Solution:

Given that,

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

$\rm \implies {x}^{2} + 6x - ( {a}^{2} + 4a - 2a - 8) = 0$

$\rm \implies {x}^{2} + 6x - (a(a + 4) - 2(a + 4)) = 0$

$\rm \implies {x}^{2} + 6x - (a- 2)(a + 4) = 0$

Now, we will split 6x.

$\rm \implies {x}^{2} + \{ (a + 4) - (a - 2)\}x - (a- 2)(a + 4) = 0$

$\rm \implies {x}^{2} + (a + 4)x - (a - 2)x - (a- 2)(a + 4) = 0$

$\rm \implies x(x +a + 4) - (a - 2) (x + a + 4)= 0$

$\rm \implies (x - a + 2) (x + a + 4)= 0$

By zero product rule,

Either x - a + 2 = 0 or x + a + 4 = 0

Therefore,

$\rm \implies x - a + 2 = 0$

$\rm \implies x = a - 2$

Again,

$\rm \implies x + a + 4 = 0$

$\rm \implies x = - a - 4$

Hence, the values of x are a - 2 and -a - 4

Answer 2

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

We need to get rid of expression parentheses.  If there is a negative sign in front of it, each term within the expression changes sign.  Otherwise, the expression remains unchanged.  In our example, the following 3 terms will change sign : a² , 2a , -8

$\to\sf{x^2+6x-a^2-2a+8=0}$

• In order to solve this non-linear equation, we need to apply the following quadratic formula :

$\bf{Quadratic\:Equation\:Formula}$

For a quadratic equation of the form ax² + bx + c = 0 the solutions are

$\sf{x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

In Our Question , We have : a = 1 , b = 6 , c = -a² - 2a + 8

$\to\sf{\displaystyle x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \left(-a^2-2a+8\right)}}{2\cdot \:1}}$

$\to\sf{\displaystyle x_{1,\:2}=\frac{-6\pm \:2\left(a+1\right)}{2\cdot \:1}}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$\sf{x=a-2,\:x=-a-4}$

$\large\boxed{\sf{x \in\{(a-2),(-a-4)\}}}$