To solve the equation [tex]\( x^4 + 95x^2 - 500 = 0 \)[/tex] using factoring, we can start by making a substitution to simplify our approach. Let's set [tex]\( y = x^2 \)[/tex]. This transforms the original equation into a quadratic equation:
[tex]\[ y^2 + 95y - 500 = 0 \][/tex]
Now, we need to solve the quadratic equation [tex]\( y^2 + 95y - 500 = 0 \)[/tex]. To do this, we can either factor it (if possible) or use the quadratic formula. We'll use the quadratic formula in this case:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( y^2 + 95y - 500 = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = 95, \quad c = -500 \][/tex]
Plugging these values into the quadratic formula gives:
[tex]\[ y = \frac{-95 \pm \sqrt{95^2 - 4 \cdot 1 \cdot (-500)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-95 \pm \sqrt{9025 + 2000}}{2} \][/tex]
[tex]\[ y = \frac{-95 \pm \sqrt{11025}}{2} \][/tex]
[tex]\[ y = \frac{-95 \pm 105}{2} \][/tex]
This results in two solutions:
[tex]\[ y = \frac{-95 + 105}{2} = \frac{10}{2} = 5 \][/tex]
[tex]\[ y = \frac{-95 - 105}{2} = \frac{-200}{2} = -100 \][/tex]
Now we revert back to our original variable [tex]\( x \)[/tex]. Recall that [tex]\( y = x^2 \)[/tex], so we now have:
[tex]\[ x^2 = 5 \quad \text{or} \quad x^2 = -100 \][/tex]
For [tex]\( x^2 = 5 \)[/tex], solving for [tex]\( x \)[/tex] gives:
[tex]\[ x = \pm \sqrt{5} \][/tex]
For [tex]\( x^2 = -100 \)[/tex], solving for [tex]\( x \)[/tex] gives:
[tex]\[ x = \pm \sqrt{-100} = \pm 10i \][/tex]
(where [tex]\( i \)[/tex] is the imaginary unit, [tex]\( i^2 = -1 \)[/tex]).
Therefore, the solutions to the equation [tex]\( x^4 + 95x^2 - 500 = 0 \)[/tex] are:
[tex]\[ x = \pm \sqrt{5} \text{ and } x = \pm 10i \][/tex]
Among the provided choices, the correct one is:
[tex]\[ x = \pm \sqrt{5} \text{ and } x = \pm 10i \][/tex]
