Question

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

Answer 1

Answer:

Step-by-step 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

$x^2=-\frac{7}{9}$

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

$x=$±$\sqrt{-\frac{7}{9} }$

which will simplify down to

$x=$±$\frac{\sqrt{7} }{3}i$

Those are the 4 solutions to the quartic equation.