Question

What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

Answer 1

Explanation:

Let
`u^2=x^4\\u = x^2`

Subbing in:

`9u^2-2u-7=0`

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

`9u^2-9u+7u-7=0`

Group together in groups of 2:

`(9u^2-9u)+(7u-7)=0`

Now factor out what's common within each set of parenthesis:

`9u(u-1)+7(u-1)=0`

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

`(u-1)(9u+7)=0`

Remember that
`u=x^2`

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

`(x^2-1)(9x^2+7)=0`

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring
`(x^2-1)` gives us that x = 1 and -1. The other set is a bit more tricky. If

`9x^2+7=0` then

`9x^2=-7` and

`x^2=-(7)/(9)`

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:

`x=`±
`\sqrt{-(7)/(9) }`

which will simplify down to

`x=`±
`(√(7) )/(3)i`

Those are the 4 solutions to the quartic equation.