Find the roots algebraically: 1st problem: x^4+4x^2=0 2nd problem: 16x^4-9x^2=0?

Question

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Answer 1

1) x^4 + 4x^2 = 0
x^2 (x^2 + 4) = 0

Do you want complex roots too? Since I don't know, I'll play it safe and assume that you do.

x^2 = 0 implies x = 0
x^2 + 4 = 0 implies x^2 = -4, which means x = +/- 2i.

Therefore, x = {0, 2i, -2i}

2) 16x^4 - 9x^2 = 0

x^2 (16x^2 - 9) = 0

Factoring as a difference of squares, we get

x^2 (4x - 3)(4x + 3) = 0

Which means x^2 = 0 (implying x = 0),
4x - 3 = 0 (implying x = 3/4)
4x + 3 = 0 (implying x = -3/4)

Therefore, x = {0, 3/4, -3/4)

Answer 2

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

x^2+4=0 no real roots.
x^2=-4
x=+/-2i

roots are 0, -2i, 2i

16x^4-9x^2=0
x^2(16x^2-9)=0
x^2=0
x-0
16x^2-9=0
(4x+3)(4x-3)=0
4x+3=0
4x=-3
x=-3/4
4x-3=0
4x=3
x=3/4
roots are 0, 3/4, -3/4

Answer 3

take x^2 comman in the first problem
so it will become x^2(x^2+4)=0
or x^2+4=0
I think the roots of this equation will be imaginary not the real one. the answer will be +2i and -2i

Answer 4

x^4+4x²=0
x²(x²+4)=0
x={0,±2i}

16x^4-9x²=0
(4x²-3x)(4x^2+3x)=0
x(4x-3)x(4x+3)=0
x²(4x-3)(4x+3)=0
x={0,±¾ }