Q:

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

Accepted Solution

A:
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 = -7The 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.