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

Question

1439 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-01T18:31:56

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

$2p^2 - 13p +20 = 0$
$rarr (p-4)(2p-5) = 0$
$rarr p=4 " or " p = 5/2$

If $p=4$
$x^2-5 = 4$
$rarr x=+3 " or " x=-3$

If $p=5/2$
$x^2-5 = 5/2$
$rarr x=+15/2 = +7 1/2 " or " x=-15/2 = -7 1/2$

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

How do you find the solution to the quadratic equation 2(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 02(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 0?