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HURRY URGENT! Which equation has exactly two real and two nonreal solutions?

A: x^4-36x^2=0
B: x^3-x^2+x-4=0
C: x^4-5x^2-36=0
D: x^3-2x^2+x-4=0

1 Answer

3 votes

Answer:

The equation which has exactly two real and non-real solutions is:

C: x^4-5x^2-36=0

Explanation:

x^4-5x^2-36=0 (given equation)

By Factorization:

x^4-9x^2+4x^2-36=0 .... (1) (Mid term break rule)

as, (-5x^2)=(-9x^2+4x^2)

and 36 =9*4

x^2(x^2 - 9) +4(x^2 - 9)=0

taking like terms common from (1)

(x^2+4) (x^2 - 9)=0

we have:

x^2 - 9=0 and x^2 + 4=0

x^2=-9 x^2= -4

Taking square root on both sides

x= (+3, -3) and x=(+2i, -2i ) (here [i=iota=[/tex] \sqrt(-1)\\[/tex])

hence,

x=(+3, -3) are real roots of given equation.

x=(+2i, -2i) are non-real roots of given equation.

User J Whitfield
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