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anonymous
 4 years ago
solve the equation using a usubstitution
(x^2+x) ^28(x^2+x)+12=0
anonymous
 4 years ago
solve the equation using a usubstitution (x^2+x) ^28(x^2+x)+12=0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Plug problem into mathway.com

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let x^2 +x be a a^2  8a + 12 = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0factor that trinomial then substitute x^2+x back for a

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2substitute u as x^+x we'll get u^28u+12=0 u^26u2u+12=0 u(u6)2(u6)=0 (u2)(u6)=0 u=2 or u=6 so x^2+x=2 and x^2+x=6 x^+x2=0 (x+2)(x1)=0 x=2 and x=1 from the other equation x^2+x6=0 (x+3)(x2)=0 x=3 and x=2 so the solution for x=3.2.1 and 2

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1First, let\[u=x^2+x\]so then we get\[u^28u+12=0 \longrightarrow (u6)(u2)=0\]so we have two possible answers\[\begin{matrix} u =6 \\ u=2 \end{matrix}\]Then we need to solve for x. Thus, if\[x^2+x=6\]then\[x^2+x6=0 \longrightarrow (x+3)(x2)=0\]so \[\begin{matrix} x=2 \\ x=3 \end{matrix}\]Otherwise, \[x^2+x=2\]so\[x^2+x2=0 \longrightarrow (x+2)(x1)=0\]so\[\begin{matrix} x= 2 \\ x=1 \end{matrix}\]Overall, then, x has four possible values.\[\begin{matrix} x={3} \\ x={2} \\ x=1 \\ x=2 \end{matrix}\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1Always glad to help.
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