Question

Solve the equation by identifying the quadratic form. Use a substitute variable(t) and find all real solutions by factoring. Type your answers from smallest to largest. If an answer is not an integer then type it as a decimal rounded to the nearest hundredth. When typing exponents do not use spaces and use the carrot key ^ (press shift and 6). For example, x cubed can be typed as x^3.(x^2-1)^2+(x^2-1)-12=0Step 1. Identify the quadratic formLet t= Answer. We now have:t^2+t-12=0Step 2. FactorFactor this and solve for t to get t=Answerand Answer Step 3. Solve for xWe have solved for t now we need to use this value for t to help us solve for x. Revisit step 1 to remind you of the relationship between t and x. Type your real solutions (no extraneous) from smallest to largest.x= Answer and Answer

Answer 1

We have the following equation

$\mleft(x^2-1\mright)^2+\mleft(x^2-1\mright)-12=0$

Step 1.

Let t = (x^2 -1)

$t=(x^2-1)$

We now have:

$t^2+t-12=0$

Step 2. factor the above, to get t = -4 and 3

$\begin{gathered} t^2+t-12=0 \\ t_{1,\: 2}=\frac{-1\pm\sqrt{1^2-4\cdot\:1\cdot\left(-12\right)}}{2\cdot\:1} \\ t_{1,\: 2}=\frac{-1\pm\sqrt{1^2-4\cdot\:1\cdot\left(-12\right)}}{2\cdot\:1} \\ t_1=\frac{-1+7}{2\cdot\:1},\: t_2=\frac{-1-7}{2\cdot\:1} \\ t_1=3,\: t_2=-4 \end{gathered}$

Step 3.

$\begin{gathered} t=(x^2-1) \\ t_1=3,\: t_2=-4 \end{gathered}$

we can't use t = -4 because we will obtain an imaginary number, instead, lets use t=3 to solve for x

$\begin{gathered} \mleft(x^2-1\mright)^{}=3 \\ x^2-1=3 \\ x^2-1+1=3+1 \\ x^2=4 \\ x=2,\: x=-2 \end{gathered}$

So, x = -2 and 2