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

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Answer 1 · 2017-02-04T05:59:40

Zeros of $y=x^4-7x^2+12$ are $-2,-sqrt3,sqrt3$ and $2$

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

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

$(u-4)(u-3)=0$

i.e. either $u=4$ i.e. $x^2=4$ or $x^2-4=0$ or $(x+2)(x-2)=0$ and $x=-2$ or $2$ .

or $u=3$ i.e. $x^2=3$ or $x^2-3=0$ or $(x+sqrt3)(x-sqrt3)=0$ and $x=-sqrt3$ or $sqrt3$ .

Hence, zeros of $y=x^4-7x^2+12$ are $-2,-sqrt3,sqrt3$ and $2$

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