How do you find the zeros, if any, of $y= -x^4 -3x^2+17$ using the quadratic formula?

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Answer 1 · 2016-02-18T22:14:23

$+-sqrt((-3+sqrt(77))/(2))$

if $a=x^2$ you will obtain:

$-a^2-3a+17$

Apply the second degree formula:

$(3+-sqrt(3^2-4xx(-1)xx17))/(2xx(-1))$

$(3+-sqrt(9+68))/(-2)$

$(3+-sqrt(77))/(-2)$

$(-3+-sqrt(77))/(2)$

We know that $x=sqrt(a)$

So the real solutions will be:

$+-sqrt((-3+sqrt(77))/(2))$

How do you find the zeros, if any, of y= -x^4 -3x^2+17 y= -x^4 -3x^2+17 using the quadratic formula?