# How do you solve 4x^4-5x^2-9=0?

May 14, 2018

$x = \setminus \pm \frac{3}{2}$

#### Explanation:

Set $t = {x}^{2}$. Then, ${t}^{2} = {\left({x}^{2}\right)}^{2} = {x}^{4}$. The equation becomes

$4 {t}^{2} - 5 t - 9 = 0$

Solve it with the usual quadratic formula to obtain ${t}_{1} = - 1$ and ${t}_{2} = \frac{9}{4}$

Now we must revert back to $x$:

${t}_{1} = - 1 \setminus \implies {x}^{2} = - 1$, which is impossible.

${t}_{2} = \frac{9}{4} \setminus \implies {x}^{2} = \frac{9}{4} \setminus \implies x = \pm \frac{3}{2}$

May 14, 2018

$x = \pm \frac{3}{2} \text{ or } x = \pm i$

#### Explanation:

$\text{make the substitution } u = {x}^{2}$

$\Rightarrow 4 {u}^{2} - 5 u - 9 = 0$

$\text{using the a-c method to factor the quadratic}$

$\text{the factors of the product } 4 \times - 9 = - 36$

$\text{which sum to - 5 are + 4 and - 9}$

$\text{split the middle term using these factors}$

$4 {u}^{2} + 4 u - 9 u - 9 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$\textcolor{red}{4 u} \left(u + 1\right) \textcolor{red}{- 9} \left(u + 1\right) = 0$

$\text{take out the "color(blue)" common factor } \left(u + 1\right)$

$\Rightarrow \left(u + 1\right) \left(\textcolor{red}{4 u - 9}\right) = 0$

$\text{equate each factor to zero and solve for u}$

$u + 1 = 0 \Rightarrow u = - 1$

$4 u - 9 = 0 \Rightarrow u = \frac{9}{4}$

$\text{change u back into terms of x}$

$\Rightarrow {x}^{2} = \frac{9}{4} \Rightarrow x = \pm \frac{3}{2} \leftarrow \textcolor{red}{\text{2 real roots}}$

${x}^{2} = - 1 \Rightarrow x = \pm i \leftarrow \textcolor{red}{\text{2 complex roots}}$