Respuesta :
Answer:
The solutions and the x-intercepts of the polynomial [tex]2x^4-24x^2+40[/tex] are:
[tex]x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}[/tex]
Step-by-step explanation:
Given a function f a solution or a root of f is a value [tex]x_{0}[/tex] at which [tex]f(x_0)=0[/tex].
An x-intercept is a point on the graph where y is zero.
To find the solutions of the polynomial and the x-intercepts [tex]2x^4-24x^2+40[/tex] you need to:
First, we need to factor the polynomial expression
Factor the common term
[tex]{\left(2 x^{4} - 24 x^{2} + 40\right)} = {\left(2 \left(x^{4} - 12 x^{2} + 20\right)\right)}[/tex]
We can treat [tex]x^{4} - 12 x^{2} + 20[/tex] as a quadratic function with respect to [tex]x^2[/tex]
Let [tex]u=x^2[/tex]. We can rewrite [tex]x^{4} - 12 x^{2} + 20[/tex] in terms of [tex]u[/tex] as follows:
[tex]u^2-12u+20[/tex]
We need to solve the quadratic equation
[tex]u^2-12u+20=0[/tex]
for this we can use the Quadratic Equation Formula:
For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\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}[/tex]
[tex]u_1=\frac{-\left(-12\right)+\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_1=10[/tex]
[tex]u_2=\frac{-\left(-12\right)-\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_2=2[/tex]
the solutions to the quadratic equation are:
[tex]u=10,\:u=2[/tex]
Therefore, [tex]u^2-12u+20=(u-10)(u-2)[/tex]
Recall that [tex]u=x^2[/tex] so
[tex]2 x^{4} - 24 x^{2} + 40=2 \left(x^{2} - 10\right) \left(x^{2} - 2\right)=0[/tex]
Using the Zero factor Theorem: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.
[tex]x^2-10=0[/tex] roots are [tex]x_1=\sqrt{10}[/tex]; [tex]x_2=-\sqrt{10}[/tex]
[tex]x^{2} - 2=0[/tex] roots are [tex]x_1=\sqrt{2}[/tex]; [tex]x_2=-\sqrt{2}[/tex]
The solutions and the x-intercepts are:
[tex]x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}[/tex]
Because all roots are real roots the x-intercepts and the solutions are equal.
