Answer:
Step-by-step explanation:
Let [tex]u^2=x^4\\u = x^2[/tex]
Subbing in:
[tex]9u^2-2u-7=0[/tex]
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:
[tex]9u^2-9u+7u-7=0[/tex]
Group together in groups of 2:
[tex](9u^2-9u)+(7u-7)=0[/tex]
Now factor out what's common within each set of parenthesis:
[tex]9u(u-1)+7(u-1)=0[/tex]
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:
[tex](u-1)(9u+7)=0[/tex]
Remember that [tex]u=x^2[/tex]
so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.
[tex](x^2-1)(9x^2+7)=0[/tex]
The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring [tex](x^2-1)[/tex] gives us that x = 1 and -1. The other set is a bit more tricky. If
[tex]9x^2+7=0[/tex] then
[tex]9x^2=-7[/tex] and
[tex]x^2=-\frac{7}{9}[/tex]
You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:
[tex]x=[/tex]±[tex]\sqrt{-\frac{7}{9} }[/tex]
which will simplify down to
[tex]x=[/tex]±[tex]\frac{\sqrt{7} }{3}i[/tex]
Those are the 4 solutions to the quartic equation.
