Math

QuestionFind the real zeros and xx-intercepts of the function f(x)=x441x2+400f(x)=x^{4}-41 x^{2}+400.

Studdy Solution

STEP 1

Assumptions1. The function is given by f(x)=x441x+400f(x)=x^{4}-41 x^{}+400 . We are looking for the real zeros of the function, which are also the x-intercepts of the graph of the function.

STEP 2

The real zeros of the function are the values of xx that make f(x)=0f(x) =0. So, we set the function equal to zero and solve for xx.
x441x2+400=0x^{4}-41 x^{2}+400 =0

STEP 3

This equation is a quadratic equation in terms of x2x^2. We can rewrite it in the form of a standard quadratic equation ax2+bx+c=0ax^2 + bx + c =0 by substituting x2=yx^2 = y.
y241y+400=0y^{2}-41 y+400 =0

STEP 4

Now, we can solve this quadratic equation by factoring. The factors of400 that add up to41 are16 and25. So, we can write the equation as(y16)(y25)=0(y -16)(y -25) =0

STEP 5

Setting each factor equal to zero gives the solutions for yyy16=0ory25=0y -16 =0 \quad \text{or} \quad y -25 =0

STEP 6

olving for yy givesy=16ory=25y =16 \quad \text{or} \quad y =25

STEP 7

Remember that y=x2y = x^2. So, we substitute x2x^2 back in for yy to find the solutions for xxx2=16orx2=25x^2 =16 \quad \text{or} \quad x^2 =25

STEP 8

Taking the square root of both sides of each equation gives the solutions for xxx=±4orx=±5x = \pm4 \quad \text{or} \quad x = \pm5The zeros and the xx-intercepts are the same. They are x=5,4,4,5x = -5, -4,4,5.

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