Question

For 15-20, solve each equation. Identifyany extraneous roots. I don’t need explanations, i’ve completed it until: x^3+6x^2+4x+24=0

Answer 1

we have the equation

$(8x-16)^{\frac{1}{3}}=x+2$

elevated to the cubic on both sides

$(8x-16)=(x+2)^3$

Expand the right side

$\begin{gathered} (8x-16)=(x+2)^2(x+2) \\ (8x-16)=(x^2+4x+4)^(x+2) \\ (8x-16)=x^3+2x^2+4x^2+8x+4x+8 \\ 8x-16=x^3+6x^2+12x+8 \\ x^3+6x^2+12x+8-8x+16=0 \\ x^3+6x^2+4x+24=0 \end{gathered}$

For the cubic equation

For x=-6

the equation is equal to zero

so

x=-6 is a root

Divide the cubic function by the factor (x+6)

x^3+6x^2+4x+24 : (x+6)

x^2+4

-x^3-6x^2

-------------------------

4x+24

-4x-24

--------------

0

therefore

x^3+6x^2+4x+24=(x+6)(x^2+4)

Solve the quadratic equation

x^2+4=0

x^2=-4

x=2i and x=-2i

the solutions are

x=-6

x=2i

x=-2i

Verify in the original equation

For x=-6

$\begin{gathered} \sqrt[3]{(8(-6)-16)}=-6+2 \\ -4=-4\text{ ----> is true} \end{gathered}$

The solution is x=-6

x=2i and x=-2i are not solutions for the given equation