How do you solve $4 x^{4} - 16 x^{2} + 15 = 0$ $4x^4 - 16x^2 + 15 = 0$ ?

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How do you solve $4 x^{4} - 16 x^{2} + 15 = 0$ $4x^4 - 16x^2 + 15 = 0$ ?

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Answer 1 · 2018-05-03T17:20:50

$\pm \sqrt{\frac{5}{2}}$ $+-sqrt(5/2)$
$\pm \sqrt{\frac{3}{2}}$ $+-sqrt(3/2)$

Explanation:

for real coefficient equation
equation of n‐th degree exist n roots
so this equations exists 3 possible answers
1. two pairs of the complex conjugate of $a + b i$ $a+bi$ & $a - b i$ $a-bi$
2. a pair of the complex conjugate of $a + b i$ $a+bi$ & $a - b i$ $a-bi$ and two real roots
3. four real roots

$4 x^{4} - 16 x^{2} + 15 = 0$ $4x^4-16x^2+15=0$
first I guess I can use "Cross method" to factorizative this equation
it can be seen as below
$(2 x^{2} - 5) (2 x^{2} - 3) = 0$ $(2x^2-5)(2x^2-3)=0$
so there are four real roots
$\pm \sqrt{\frac{5}{2}}$ $+-sqrt(5/2)$
$\pm \sqrt{\frac{3}{2}}$ $+-sqrt(3/2)$

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How do you solve $4 x^{4} - 16 x^{2} + 15 = 0$ $4x^4 - 16x^2 + 15 = 0$ ?

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How do you solve $4 x^{4} - 16 x^{2} + 15 = 0$ $4x^4 - 16x^2 + 15 = 0$ ?