Solve the quadratic equation $ 2{x^2} + ax - {a^2} = 0 $ for x

Hint: So by using the factorization method, we can find the solutions of $ 2{x^2} + ax - {a^2} = 0 $ . Tip for factoring, we have to divide the middle term in such a way that the product of the coefficients of the divided terms must be equal to the product of the coefficients of the first and third terms. The solutions of x must be in terms of a.

Complete step-by-step answer:
We are given to solve the quadratic equation $ 2{x^2} + ax - {a^2} = 0 $
Here the product of the coefficients of the first and last terms is $ 2 \times 1 = 2 $ . So the middle term ax must be divided in such a way that the product of the coefficients of the divided terms must be equal to 2.
So ‘ax’ can be divided as $ 2ax - ax $ , as $ 2ax - ax = ax $ and the product of coefficients of 2ax and ax is $ 2 \times 1 = 2 $ .
So $ 2{x^2} + ax - {a^2} = 0 $ can also be written as $ 2{x^2} + 2ax - ax - {a^2} = 0 $
From the first two terms of the above equation, taking out ‘2x’ common and from the last two terms taking out ‘a’ common, we get
 $ \Rightarrow 2x\left( {x + a} \right) - a\left( {x + a} \right) = 0 $
 $ \Rightarrow \left( {x + a} \right)\left( {2x - a} \right) = 0 $
The left hand side of the above equation is a product which is equal to zero. So either the first term or the second term or both the terms must be zero.
This gives, $ \Rightarrow \left( {x + a} \right) = 0,\left( {2x - a} \right) = 0 $
 $ \Rightarrow x = - a,x = \dfrac{a}{2} $
The values of x are $ - a,\dfrac{a}{2} $ .
So, the correct answer is “ $ - a,\dfrac{a}{2} $ ”.

Note: Be careful while dividing the middle term or the term with ‘x’ while factoring. No. of solutions of an equation can be determined by the highest degree of the variable of the equation. Every quadratic equation will have exactly 2 factors as its highest degree is 2.