Answer :
To solve the equation [tex]\( x^4 + 95 x^2 - 500 = 0 \)[/tex] using factoring, we can try to transform it into a quadratic form. Notice that the equation involves [tex]\( x^4 \)[/tex] and [tex]\( x^2 \)[/tex]. Let’s introduce a substitution to simplify the process:
Let [tex]\( u = x^2 \)[/tex]. Therefore, [tex]\( x^4 = u^2 \)[/tex]. Substituting these into the original equation, we get:
[tex]\[ u^2 + 95u - 500 = 0. \][/tex]
This is now a quadratic equation in terms of [tex]\( u \)[/tex]. To solve this quadratic equation, we can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 95 \)[/tex], and [tex]\( c = -500 \)[/tex]. Plugging in these values, we obtain:
[tex]\[ u = \frac{-95 \pm \sqrt{95^2 - 4 \cdot 1 \cdot (-500)}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-95 \pm \sqrt{9025 + 2000}}{2} \][/tex]
[tex]\[ u = \frac{-95 \pm \sqrt{11025}}{2} \][/tex]
[tex]\[ u = \frac{-95 \pm 105}{2} \][/tex]
This results in two possible solutions for [tex]\( u \)[/tex]:
[tex]\[ u_1 = \frac{-95 + 105}{2} = \frac{10}{2} = 5 \][/tex]
[tex]\[ u_2 = \frac{-95 - 105}{2} = \frac{-200}{2} = -100 \][/tex]
Now, we revert back to [tex]\( x \)[/tex] using our substitution [tex]\( u = x^2 \)[/tex]:
1. For [tex]\( u_1 = 5 \)[/tex]:
[tex]\[ x^2 = 5 \][/tex]
[tex]\[ x = \pm \sqrt{5} \][/tex]
2. For [tex]\( u_2 = -100 \)[/tex]:
[tex]\[ x^2 = -100 \][/tex]
[tex]\[ x = \pm \sqrt{-100} \][/tex]
[tex]\[ x = \pm 10i \][/tex]
Therefore, the solutions to the equation [tex]\( x^4 + 95 x^2 - 500 = 0 \)[/tex] are:
[tex]\[ \boxed{x = \pm \sqrt{5} \text{ and } x = \pm 10i} \][/tex]
From the given options, the correct answer corresponds to:
[tex]\[ x= \pm \sqrt{5} \text{ and } x= \pm 10 i \][/tex]
Let [tex]\( u = x^2 \)[/tex]. Therefore, [tex]\( x^4 = u^2 \)[/tex]. Substituting these into the original equation, we get:
[tex]\[ u^2 + 95u - 500 = 0. \][/tex]
This is now a quadratic equation in terms of [tex]\( u \)[/tex]. To solve this quadratic equation, we can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 95 \)[/tex], and [tex]\( c = -500 \)[/tex]. Plugging in these values, we obtain:
[tex]\[ u = \frac{-95 \pm \sqrt{95^2 - 4 \cdot 1 \cdot (-500)}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-95 \pm \sqrt{9025 + 2000}}{2} \][/tex]
[tex]\[ u = \frac{-95 \pm \sqrt{11025}}{2} \][/tex]
[tex]\[ u = \frac{-95 \pm 105}{2} \][/tex]
This results in two possible solutions for [tex]\( u \)[/tex]:
[tex]\[ u_1 = \frac{-95 + 105}{2} = \frac{10}{2} = 5 \][/tex]
[tex]\[ u_2 = \frac{-95 - 105}{2} = \frac{-200}{2} = -100 \][/tex]
Now, we revert back to [tex]\( x \)[/tex] using our substitution [tex]\( u = x^2 \)[/tex]:
1. For [tex]\( u_1 = 5 \)[/tex]:
[tex]\[ x^2 = 5 \][/tex]
[tex]\[ x = \pm \sqrt{5} \][/tex]
2. For [tex]\( u_2 = -100 \)[/tex]:
[tex]\[ x^2 = -100 \][/tex]
[tex]\[ x = \pm \sqrt{-100} \][/tex]
[tex]\[ x = \pm 10i \][/tex]
Therefore, the solutions to the equation [tex]\( x^4 + 95 x^2 - 500 = 0 \)[/tex] are:
[tex]\[ \boxed{x = \pm \sqrt{5} \text{ and } x = \pm 10i} \][/tex]
From the given options, the correct answer corresponds to:
[tex]\[ x= \pm \sqrt{5} \text{ and } x= \pm 10 i \][/tex]
