How do you find the solution to the quadratic equation $2 {(x^{2} - 5)}^{2} - 13 (x^{2} - 5) + 20 = 0$ $2(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 0$ ?

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How do you find the solution to the quadratic equation $2 {(x^{2} - 5)}^{2} - 13 (x^{2} - 5) + 20 = 0$ $2(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 0$ ?

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Answer 1 · 2015-05-01T18:31:56

Let $p = (x^{2} - 5)$ $p=(x^2-5)$
and solve
$2 p^{2} - 13 p + 20 = 0$ $2p^2 - 13p +20 = 0$
Then for each solution (in terms of $p$ $p$ )
solve for $x$ $x$

$2 p^{2} - 13 p + 20 = 0$ $2p^2 - 13p +20 = 0$
$\to (p - 4) (2 p - 5) = 0$ $rarr (p-4)(2p-5) = 0$
$\to p = 4 or p = \frac{5}{2}$ $rarr p=4 " or " p = 5/2$

If $p = 4$ $p=4$
$x^{2} - 5 = 4$ $x^2-5 = 4$
$\to x = + 3 or x = - 3$ $rarr x=+3 " or " x=-3$

If $p = \frac{5}{2}$ $p=5/2$
$x^{2} - 5 = \frac{5}{2}$ $x^2-5 = 5/2$
$\to x = + \frac{15}{2} = + 7 \frac{1}{2} or x = - \frac{15}{2} = - 7 \frac{1}{2}$ $rarr x=+15/2 = +7 1/2 " or " x=-15/2 = -7 1/2$

So
$x ε {- 7 \frac{1}{2}, - 3, + 3, + 7 \frac{1}{2}}$ $x epsilon {-7 1/2, -3, +3, +7 1/2}$

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How do you find the solution to the quadratic equation $2 {(x^{2} - 5)}^{2} - 13 (x^{2} - 5) + 20 = 0$ $2(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 0$ ?

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How do you find the solution to the quadratic equation $2 {(x^{2} - 5)}^{2} - 13 (x^{2} - 5) + 20 = 0$ $2(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 0$ ?