Question

What situation should be used to rewrite 16(x^3+1)^2 -22(x^3+1)-3=0 as a quadratic equation

Answer 1

`16(x^3+1)^2-22(x^3+1)-3=0`

Use substitution:
`x^3+1=t`


`16t^2-22t-3=0\\\\16t^2-24t+2t-3=0\\\\8t(2t-3)+1(2t-3)=0\\\\(2t-3)(8t+1)=0\iff2t-3=0\ \vee\ 8t+1=0`


`2t-3=0\ \ \ |+3\\\\2t=3\ \ \ |:2\\\\t=(3)/(2)\\..............................\\8t+1=0\ \ \ \ |-1\\\\8t=-1\ \ \ \ |:8\\\\t=-(1)/(8)`

we're going back to substitution:


`x^3+1=(3)/(2)\ \vee\ x^3+1=-(1)/(8)\ \ \ \ |subtract\ 1\ from\ both\ sides\ of\ the\ equations\\\\x^3=(1)/(2)\ \vee\ x^3=-(9)/(8)\\\\x=\sqrt[3]{(1)/(2)}\ \vee\ x=\sqrt[3]{-(9)/(8)}`



`x=(1)/(\sqrt[3]2)\ \vee\ x=-(\sqrt[3]9)/(2)\\\\\boxed{x=(\sqrt[3]4)/(2)\ \vee\ x=-(\sqrt[3]9)/(2)}`