Question

Find the solutions to 2x⁴-24x²+40=0 and the x-intercepts of the graph of y=2x⁴-24x²+40.

Answer 1

The solutions and the x-intercepts of the polynomial
`2x^4-24x^2+40` are:

`x=√(10),\:x=-√(10),\:x=√(2),\:x=-√(2)`

Explanation:

Given a function f a solution or a root of f is a value
`x_(0)` at which
`f(x_0)=0`.

An x-intercept is a point on the graph where y is zero.

To find the solutions of the polynomial and the x-intercepts
`2x^4-24x^2+40` you need to:

First, we need to factor the polynomial expression

Factor the common term

`{\left(2 x^(4) - 24 x^(2) + 40\right)} = {\left(2 \left(x^(4) - 12 x^(2) + 20\right)\right)}`

We can treat
`x^(4) - 12 x^(2) + 20` as a quadratic function with respect to
`x^2`

Let
`u=x^2`. We can rewrite
`x^(4) - 12 x^(2) + 20` in terms of
`u` as follows:

`u^2-12u+20`

We need to solve the quadratic equation

`u^2-12u+20=0`

for this we can use the Quadratic Equation Formula:

For a quadratic equation of the form
`ax^2+bx+c=0` the solutions are

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=1,\:b=-12,\:c=20:\quad u_(1,\:2)=(-\left(-12\right)\pm √(\left(-12\right)^2-4\cdot \:1\cdot \:20))/(2\cdot \:1)`

`u_1=(-\left(-12\right)+√(\left(-12\right)^2-4\cdot \:1\cdot \:20))/(2\cdot \:1)\\u_1=10`

`u_2=(-\left(-12\right)-√(\left(-12\right)^2-4\cdot \:1\cdot \:20))/(2\cdot \:1)\\u_2=2`

the solutions to the quadratic equation are:

`u=10,\:u=2`

Therefore,
`u^2-12u+20=(u-10)(u-2)`

Recall that
`u=x^2` so

`2 x^(4) - 24 x^(2) + 40=2 \left(x^(2) - 10\right) \left(x^(2) - 2\right)=0`

Using the Zero factor Theorem: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.

`x^2-10=0` roots are
`x_1=√(10)`;
`x_2=-√(10)`

`x^(2) - 2=0` roots are
`x_1=√(2)`;
`x_2=-√(2)`

The solutions and the x-intercepts are:

`x=√(10),\:x=-√(10),\:x=√(2),\:x=-√(2)`

Because all roots are real roots the x-intercepts and the solutions are equal.