# How do you find the zeros of y=x^4-7x^2+12?

Feb 4, 2017

Zeros of $y = {x}^{4} - 7 {x}^{2} + 12$ are $- 2 , - \sqrt{3} , \sqrt{3}$ and $2$

#### Explanation:

Zeroa of $y = {x}^{4} - 7 {x}^{2} + 12$ will be given by the solution to the equation ${x}^{4} - 7 {x}^{2} + 12 = 0$. To solve this assume $u = {x}^{2}$, then the equation becomes

${u}^{2} - 7 u + 12 = 0$, which can be factorized as

$\left(u - 4\right) \left(u - 3\right) = 0$

i.e. either $u = 4$ i.e. ${x}^{2} = 4$ or ${x}^{2} - 4 = 0$ or $\left(x + 2\right) \left(x - 2\right) = 0$ and $x = - 2$ or $2$.

or $u = 3$ i.e. ${x}^{2} = 3$ or ${x}^{2} - 3 = 0$ or $\left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right) = 0$ and $x = - \sqrt{3}$ or $\sqrt{3}$.

Hence, zeros of $y = {x}^{4} - 7 {x}^{2} + 12$ are $- 2 , - \sqrt{3} , \sqrt{3}$ and $2$