Question

Consider the polynomial P(x)=x⁴-3x²-4.How many x-intercepts does the graph of the equation y=x⁴-3x²-4 have? What are the x-intercepts?

Answer 1

The equation has two x-intercepts. They are
`x = 2` and
`x = -2`.

Explanation:

A x-intercept are the values of x when
`y = 0`.

The number of x-intercepts is the number of solutions. The x-intercepts are the solutions.

So let's solve the equation:

We have the following equation:

`1)x^(4) - 3x^(2) - 4 = 0`.

The first step to solve this problem is using

`2) y = x^(2)`

We replace in the equation 1, find the values of y, and then we replace in equation 2) to find the values of x.

To solve the equations, it is important to know how we find the roots of a second order polynomial.

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = (x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

In this problem, we have

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

`y = x^(2)`

So

`y^(2) - 3y - 4 = 0`

So:
`a = 1, b = -3, c = -4`

`\bigtriangleup = b^(2) - 4ac = (-3)^(2) -4(1)(-4) = 25`

`y_(1) = (-b + √(\bigtriangleup))/(2*a) = (3 + √(25))/(2) = 4`

`y_(2) = (-b - √(\bigtriangleup))/(2*a) = (3 - √(25))/(2) = -1`

The values of y are
`y_(1) = 4, y_(2) = -1`

We also have that:

`y = x^(2)`

So

`4 = x^(2)`

`x = \pm √(4)`

`x = \pm 2`

And

`-1 = x^(2)`

There is no real solution for this. So our only solutions are
`x = 2` and
`x = -2`.

So, the equation has two x-intercepts. They are
`x = 2` and
`x = -2`.