Question

What are the solutions of the equation x4 – 5x2 – 14 = 0? Use factoring to solve.

Answer 1

Answer:  The required solutions of the given equation are

$x=\pm\sqrt7,~\pm\sqrt2i.$

Step-by-step explanation:  We are given to find the solutions of the following bi-quadratic equation :

$x^4-5x^2-14=0~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$x^2=y~~~~~\Rightarrow x^4=y^2.$

So, equation (i) becomes

$y^2-5y-14=0,$

which is a quadratic equation in the variable y. The above equation in variable y can be solved using factorization as follows :

$y^2-5y-14=0\\\\\Rightarrow y^2-7y+2y-14=0\\\\\Rightarrow y(y-7)+2(y-7)=0\\\\\Rightarrow (y-7)(y+2)=0\\\\\Rightarrow y-7=0,~~~y+2=0\\\\\Rightarrow y=7,-2.$

Therefore, we get

$x^2=7\\\\\Rightarrow x=\pm\sqrt7$

and

$x^2=-2\\\\\Rightarrow x=\pm\sqrt{-2}\\\\\Rightarrow x=\pm\sqrt2i,~~~~~~~~~~~~~[\textup{where }i=\sqrt{-1}]$

Thus, the required solutions of the given equation are

$x=\pm\sqrt7,~\pm\sqrt2i.$

Answer 2

(x2 - ) (x2 + ) = 0
(x2 - 7) (x2 + 2) = 0
(x2 - 7) = 0 and (x2 + 2) = 0
x2 = 7 and x2 = - 2
√x2 = √7 and √x2 = √-2
x = √7 and x = √-2 --> it doesn't exist