How do you solve quadratic equations to the fourth power?

Answer: guess roots and factor.

Quadratic equations are polynomial equations whose highest power is 2, while quartic equations are polynomial equations whose highest power is 4.

Any polynomial expression (whether quadratic, quartic, or any other power) can be factored into its roots. A root is a value which, when plugged in for the variable, equals zero.

Example: ##x^4 + 3x^3 + x^2 – x + 2 = 2x + 4##.

First rewrite as a polynomial that equals 0: ##x^4 + 3x^3 + x^2 -3x -2 = 0##.

A quartic equation has at most 4 roots. It isn’t elegant, but we can find them by guessing a value for ##x## and seeing whether it equals zero. 1 works here because ##1^4 + 3*1^3 + 1^2 – 3*1 – 2 = 0##.

Now that we have one root, 1, we know that the expression can be factored into ##(x-1)## and a cubic because plugging 1 into the ##(x-1)## term equals 0.

To find the cubic, we use polynomial long division:

##(x^4+3x^3+x^2-3x-2)/(x-1) = x^3 + 4x^2 + 5x + 2##

Continuing, we can find the other roots -1 (a double root) and -2. Or you can use the cubic or quadratic equation once you’ve reduced the polynomial.

This gives the final factorization of ##(x-1)(x+1)^2(x+2) = 0##. So the solutions are ##x = 1, -1, text{or} -2##.

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How do you solve quadratic equations to the fourth power?

Answer: guess roots and factor.

Quadratic equations are polynomial equations whose highest power is 2, while quartic equations are polynomial equations whose highest power is 4.

Any polynomial expression (whether quadratic, quartic, or any other power) can be factored into its roots. A root is a value which, when plugged in for the variable, equals zero.

Example: ##x^4 + 3x^3 + x^2 – x + 2 = 2x + 4##.

First rewrite as a polynomial that equals 0: ##x^4 + 3x^3 + x^2 -3x -2 = 0##.

A quartic equation has at most 4 roots. It isn’t elegant, but we can find them by guessing a value for ##x## and seeing whether it equals zero. 1 works here because ##1^4 + 3*1^3 + 1^2 – 3*1 – 2 = 0##.

Now that we have one root, 1, we know that the expression can be factored into ##(x-1)## and a cubic because plugging 1 into the ##(x-1)## term equals 0.

To find the cubic, we use polynomial long division:

##(x^4+3x^3+x^2-3x-2)/(x-1) = x^3 + 4x^2 + 5x + 2##

Continuing, we can find the other roots -1 (a double root) and -2. Or you can use the cubic or quadratic equation once you’ve reduced the polynomial.

This gives the final factorization of ##(x-1)(x+1)^2(x+2) = 0##. So the solutions are ##x = 1, -1, text{or} -2##.


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