Question

solve the equation using a u-substitution
(x^2+x) ^2-8(x^2+x)+12=0

Answer 1

Plug problem into mathway.com

Answer 2

Let x^2 +x be a

a^2 - 8a + 12 = 0

Answer 3

factor that trinomial then substitute x^2+x back for a

Answer 4

substitute u as x^+x
we'll get
u^2-8u+12=0
u^2-6u-2u+12=0
u(u-6)-2(u-6)=0
(u-2)(u-6)=0
u=2 or u=6
so
x^2+x=2 and x^2+x=6
x^+x-2=0
(x+2)(x-1)-=0
x=-2 and x=1
from the other equation
x^2+x-6=0
(x+3)(x-2)=0
x=-3 and
x=2
so
the solution for x=-3.-2.1 and 2

Answer 5

First, let\[u=x^2+x\]so then we get\[u^2-8u+12=0 \longrightarrow (u-6)(u-2)=0\]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\[x^2+x-6=0 \longrightarrow (x+3)(x-2)=0\]so \[\begin{matrix} x=2 \\ x=-3 \end{matrix}\]Otherwise, \[x^2+x=2\]so\[x^2+x-2=0 \longrightarrow (x+2)(x-1)=0\]so\[\begin{matrix} x= -2 \\ x=1 \end{matrix}\]Overall, then, x has four possible values.\[\begin{matrix} x={-3} \\ x={-2} \\ x=1 \\ x=2 \end{matrix}\]

Answer 6

thanks once again!!!

Answer 7

Always glad to help.