Question

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

Answer 1

Answer:

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

$x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}$

Step-by-step 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

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$u_1=\frac{-\left(-12\right)+\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_1=10$

$u_2=\frac{-\left(-12\right)-\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_2=2$

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 $x_1=\sqrt{10}$; $x_2=-\sqrt{10}$

$x^{2} - 2=0$ roots are $x_1=\sqrt{2}$; $x_2=-\sqrt{2}$

The solutions and the x-intercepts are:

$x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}$

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