Question

What are the Roots/Zeroes/X-Intercepts of the following equation:


`Y =-8x^(4) +3x^(2) +9`



Please help, what process do I apply ?

Answer 1

`x = \sqrt{(3 + 3√(33))/(16)}` or

`x = -\sqrt{(3 + 3√(33))/(16)}` or

`x = i\sqrt{(3 - 3√(33))/(16)}` or

`x = -i\sqrt{(3 - 3√(33))/(16)}`

Explanation:

You can use a substitution and treat it as a quadratic equation.

Let u = x^2.

y = -8x^4 + 3x² + 9

Apply the substitution.

y = -8u² + 3u + 9

-8u² + 3u + 9 = 0

8u² - 3u - 9 = 0

Use the quadratic formula to solve for u.

`u = (-(-3) \pm √((-3)^2 - 4(8)(-9)))/(2(8))`

`u = (3 \pm √(9 + 288))/(16)`

`u = (3 \pm √(297))/(16)`

`u = (3 \pm 3√(33))/(16)`

`u = (3 + 3√(33))/(16)` or
`u = (3 - 3√(33))/(16)`

Remember that u = x², so now we substitute back x² for u.

`x^2 = (3 + 3√(33))/(16)` or
`x^2 = (3 - 3√(33))/(16)`

`x = \sqrt{(3 + 3√(33))/(16)}` or

`x = -\sqrt{(3 + 3√(33))/(16)}` or

`x = \sqrt{(3 - 3√(33))/(16)}` or

`x = -\sqrt{(3 - 3√(33))/(16)}`

Two of those roots are complex, so we deal with that now.

`x = \sqrt{(3 + 3√(33))/(16)}` or

`x = -\sqrt{(3 + 3√(33))/(16)}` or

`x = i\sqrt{(3 - 3√(33))/(16)}` or

`x = -i\sqrt{(3 - 3√(33))/(16)}`