# What are all the possible rational zeros for f(x)=2x^4-9x^2+7?

Nov 5, 2016

The rational roots are $x = \pm 1$.

#### Explanation:

Let $a$ represent ${x}^{2}$, thus $f \left(x\right) = 2 {x}^{4} - 9 {x}^{2} + 7$ can be rewritten as $2 {a}^{2} - 9 a + 7$.

This is simply a matter of factoring now, remember, to find the zeroes, you must set the equation equal to zero:

$2 {a}^{2} - 9 a + 7 = 0$

(Note that a was used rather than x, as a is the variable used in the quadratic equation, we are finding the roots of a)

$a = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$= \frac{- \left(- 9\right) \pm \sqrt{{\left(- 9\right)}^{2} - 4 \left(2\right) \left(7\right)}}{2 \left(2\right)}$
$= \frac{9 \pm \sqrt{81 - 56}}{4}$
$= \frac{9 \pm \sqrt{25}}{4}$
$= \frac{9 \pm 5}{4}$

$\therefore {a}_{1} = \frac{9 + 5}{4} = \frac{7}{2}$
$\therefore {a}_{2} = \frac{9 - 5}{4} = 1$

Recall $a = {x}^{2}$

Thus,
$\therefore {x}_{1}^{2} = \frac{7}{2}$
$\therefore {x}_{1} = \pm \sqrt{\frac{7}{2}} = \pm \frac{\sqrt{14}}{2}$

$\therefore {x}_{2}^{2} = 1$
$\therefore {x}_{2} = \pm 1$

Therefore, as the question is asking for rational roots only, $x = \pm 1.$

Hope this helped, formatting took so long :P