Question

For 15-20, solve each equation. Identifyany extraneous roots. I’ve completed most of the problem, I just am struggling to find ‘x’

Answer 1

Solutions: x = -6, x = 2i, and x = -2i

Step-by-step explanation

To solve this equation, first, we have to raise both sides of the equation to the exponent 3,

`\begin{gathered} \left({(8x-16)^(1\/3))^3}\right.=(x+2)^3 \\ \\ 8x-16=(x+2)^3 \end{gathered}`

Now, we have to expand the binomial cubed using the cube of a binomial rule,

`(a+b)^3=a^3+3a^2b+3ab^2+b^3`

So we have,

`(x+2)^3=x^3+3x^2\cdot2+3x\cdot2^2+2^3=x^3+6x^2+12x+8`

So the equation is,

`8x-16=x^3+6x^2+12x+8`

Subtract 8x from both sides of the equation, and also add 16 to both sides,

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

So, we have to find the zeros of this equation.

First, assuming at least one root is an integer, write the possible values. Since the constant is 24, the possible roots are factors of 24:

Now, let's test these values in the equation,

`\begin{gathered} f(2)=64;f(-2)=32 \\ f(3)=117;f(-3)=39 \\ ... \\ f(-6)=0 \\ ... \end{gathered}`

*Note: I have not tested all the possible values in this answer to make the explanation shorter.

Now, we know that x = -6 is a zero, so (x + 6) is a factor. To find the quadratic factor, we can divide the 3rd-degree polynomial by the factor (x + 6):

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

The zeros of the resulting factor are not real,

`\begin{gathered} x^2+4=0 \\ x=√(-4)=\pm2i \end{gathered}`

Hence, the solutions of this equation are x = -6, x = 2i, and x = -2i.