Question

Solve the following Quadratic equation:
 \[{x^2} + 6x - \left( {{a^2} + 2a - 8} \right) = 0\]

Answer 1

Hint: Here, a constant term is given in the form of quadratic polynomial, so factorize the constant term in the given quadratic equation. Then solve the given quadratic equation using the factorization method. Find the value of x in terms of a.

Complete step-by-step answer:
We have, \[{x^2} + 6x - \left( {{a^2} + 2a - 8} \right) = 0\]
Consider \[{a^2} + 2a - 8\]
This is a quadratic polynomial.
Splitting middle term, putting 2a = 4a – 2a
 \[ \Rightarrow {a^2} + 4a - 2a - 8\]
 $ a(a + 4) – 2(a + 4) $
Taking (a + 4) common
 $ (a + 4)(a – 2) $
Quadratic equation: \[{x^2} + 6x - \left( {{a^2} + 2a - 8} \right) = 0\] ⇒ \[{x^2} + 6x - \left( {a + 4} \right)(a - 2) = 0\]
Splitting middle term, 6x = (a + 4)x – (a −2)x
 \[ \Rightarrow {x^2} + \left( {a + 4} \right)x-\left( {a - 2} \right)x - \left( {a + 4} \right)(a - 2) = 0\]
 \[ \Rightarrow x\left( {x + \left( {a + 4} \right)} \right)-\left( {a - 2} \right)\left( {x + \left( {a + 4} \right)} \right) = 0\]
Taking (x + (a + 4) common
 \[ \Rightarrow \left( {x + \left( {a + 4} \right)} \right)\left( {x-\left( {a - 2} \right)} \right) = 0\]
Either $ (x + (a + 4)) = 0 $ or $ (x – (a – 2)) = 0 $
 $ x = − (a + 4) or x = a – 2 $
So, the correct answer is “x = − (a + 4) or x = a – 2”.

Note: Factorization method is preferred for these types of questions as it does not require so much calculation. But be careful while splitting middle term, product of middle term should be exactly equal to constant term. Find the values of x in terms of given constant, here a is considered as constant.