Answer:
* x ⇒ plus-or-minus StartRoot 5 EndRoot and ±10i *
Step-by-step explanation:
Through factorization;
x^4 + 95x^2 – 500 = 0, ⇒ u^2 ⇒ x^2,
u^2 + 95u - 500 = 0, ⇒ factors of -500; -25, 20, -50 * 10, -2 * 250, -5 * 100...
( u )^2 + ( - 5 + 100 ) + ( - 5 * 100 ) = 0,
( u - 5 )( u + 100 ) = 0,
u = 5, and y = - 100
Substitute u back into format x^2;
u = x^2, 5 = x^2, | x | = √5 ⇒ x = - / + √5,
u = x^2, - 100 = x^2, | x | = √- 100 ⇒ x = - / + 10i
* Answer: x ⇒ plus-or-minus StartRoot 5 EndRoot and ±10i *
A equation in form of [tex]ax^{2} +bx+c=0[/tex] is known as quadratic equation.
Solutions of given equations are [tex]x=\pm\sqrt{5}[/tex] and [tex]x=\pm10i[/tex]
Given equation are, [tex]x^{4} +95x^{2} -500=0[/tex]
Substitute [tex]x^{2} =m[/tex] in above equations. So, equation become
[tex]m^{2}+95m-500=0\\\\m^{2}+100m-5m-500=0\\\\m(m+100)-5(m+100)=0\\\\(m-5)(m+100)=0\\\\m=5,m=-100[/tex]
So, [tex]x^{2} =5,x=\pm\sqrt{5}[/tex]
[tex]x^{2} =-100,x=\pm10i[/tex]
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