Question

Find the solutions to x⁴-5x²-36=0 and the x-intercepts of the graph of y=x⁴-5x²-36.

Answer 1

The solutions
`x^4-5x^2-36=0` are
`x=3,\:x=-3,\:x=2i,\:x=-2i` and the x-intercepts of
`y=x^4-5x^2-36` are
`x=3,\:x=-3`

Explanation:

Finding the solutions to
`x^4-5x^2-36` means finding the roots, a root is where the function is equal to zero.

The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero.

To find the roots you need to:

Rewrite the equation with
`u=x^2` and
`u^2=x^4`

`u^2-5u-36=0`

Solve by factoring

`\mathrm{Break\:the\:expression\:into\:groups}\\u^2-5u-36=\left(u^2+4u\right)+\left(-9u-36\right)`

`\mathrm{Factor\:out\:}u\mathrm{\:from\:}u^2+4u=u\left(u+4\right)`

`\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9u-36=-9\left(u+4\right)`

`u^2-5u-36=u\left(u+4\right)-9\left(u+4\right)`

`\mathrm{Factor\:out\:common\:term\:}u+4\\\left(u+4\right)\left(u-9\right)`

`u^2-5u-36=\left(u+4\right)\left(u-9\right)=0`

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions to the quadratic equation are:

`\:u=-4,\:u=9`

Substitute back
`u=x^2`, solve for x

`x^4-5x^2-36=(u-9)(u+4)=(x^2-9)(x^2+4)`

Apply the difference of squares formula

`x^4-5x^2-36=(x^2-9)(x^2+4)=(x-3)(x+3)(x^2+4)`

`(x-3)(x+3)(x^2+4)=0`

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions are:

`\:x=3,\:x=-3,\:x=2i,\:x=-2i`

Because two of the solutions are complex roots the only x-intercepts are
`x=3,\:x=-3`