How do you solve $2 x^{4} - 128 = 0$ $2x^4-128=0$ ?

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How do you solve $2 x^{4} - 128 = 0$ $2x^4-128=0$ ?

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Answer 1 · 2017-01-10T23:51:45

The roots of this quartic equation are:

$\pm 2 \sqrt{2}$ $+-2sqrt(2)" "$ and $\pm 2 \sqrt{2} i$ $" "+-2sqrt(2)i$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

So we find:

$0 = 2 x^{4} - 128$ $0 = 2x^4-128$

$0 = 2 (x^{4} - 64)$ $color(white)(0) = 2(x^4-64)$

$0 = 2 ({(x^{2})}^{2} - 8^{2})$ $color(white)(0) = 2((x^2)^2-8^2)$

$0 = 2 (x^{2} - 8) (x^{2} + 8)$ $color(white)(0) = 2(x^2-8)(x^2+8)$

$0 = 2 (x^{2} - {(2 \sqrt{2})}^{2}) (x^{2} - {(2 \sqrt{2} i)}^{2})$ $color(white)(0) = 2(x^2-(2sqrt(2))^2)(x^2-(2sqrt(2)i)^2)$

$0 = 2 (x - 2 \sqrt{2}) (x + 2 \sqrt{2}) (x - 2 \sqrt{2} i) (x + 2 \sqrt{2} i)$ $color(white)(0) = 2(x-2sqrt(2))(x+2sqrt(2))(x-2sqrt(2)i)(x+2sqrt(2)i)$

So the roots of the quartic equation are:

$\pm 2 \sqrt{2}$ $+-2sqrt(2)" "$ and $\pm 2 \sqrt{2} i$ $" "+-2sqrt(2)i$

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How do you solve $2 x^{4} - 128 = 0$ $2x^4-128=0$ ?

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Explanation:

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How do you solve 2x4−128=02x^4-128=0?

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How do you solve $2 x^{4} - 128 = 0$ $2x^4-128=0$ ?