Question

need to find 2 solutions to this equation

Answer 1

\[(3x+8)^{2}-4(3x+8)-77=0\]

Answer 2

Let k = (3x+8)
So the above equation becomes
\[k^2 - 4k - 77 = 0\]

Solve the quadratic for two solutions of k, then replace k in those two equations with 3x+8 and solve for your two x solutions.

Answer 3

ok thanks, I will try that out. Not confident I'll get it right so you can correct me if it's wrong :)

Answer 4

i'm not sure if i'm doing this step correct. I got -76 for one..?

Answer 5

\[k = \frac{4 \pm \sqrt{16 + 4*77}}{2}\]
\[k = 2 \pm \sqrt{324}\]
\[k = 2 \pm 18\]
\[k = -16 \text{ or } k = 20\]
Therefore
\[3x+8 = -16\]
or
\[3x+8 = 20\]

Answer 6

So working with the first equation:
\[3x + 8 = -16\]
\[3x = -24\]
\[x = -8\]

Answer 7

Working the second one:
\[3x+8 = 20\]
\[3x = 12\]
\[x = 4\]

Answer 8

So our two solutions are \(x \in \{-8,4\}\)