Question

determine which values listed are solutions to the equation $3x^2 - 27x - 30$.
$x = 3$, $x = -1$ and $x = 10$.
justify your answer.

Explanation:

Step1: Set equation to 0

$$3x^2 - 27x - 30 = 0$$

Step2: Simplify the equation

Divide all terms by 3:
$$x^2 - 9x - 10 = 0$$

Step3: Factor the quadratic

Find two numbers that multiply to -10 and add to -9:
$$(x - 10)(x + 1) = 0$$

Step4: Solve for x

Set each factor equal to 0:
$x - 10 = 0 \implies x = 10$
$x + 1 = 0 \implies x = -1$

Step5: Verify test values

For $x=3$: $3(3)^2 -27(3)-30=27-81-30=-84
eq0$, so not a solution.
For $x=-1$: $3(-1)^2 -27(-1)-30=3+27-30=0$, so it is a solution.
For $x=10$: $3(10)^2 -27(10)-30=300-270-30=0$, so it is a solution.

Answer:

The solutions are $x=-1$ and $x=10$. $x=3$ is not a solution.