given the equation $5x^2 - 90x + 400 = 0$, determine if $x = 0$, $x = 10$ and $x = -8$ are solutions. justify your answer.

Question

Accepted Answer

# Explanation:
## Step1: Check $ x = 0 $
Substitute $ x = 0 $ into the equation $ 5x^2 - 90x + 400 = 0 $.
$$
\begin{align*}
5(0)^2 - 90(0) + 400 &= 0 - 0 + 400\
&= 400
\end{align*}
$$
Since $ 400
eq 0 $, $ x = 0 $ is not a solution.

## Step2: Check $ x = 10 $
Substitute $ x = 10 $ into the equation $ 5x^2 - 90x + 400 = 0 $.
$$
\begin{align*}
5(10)^2 - 90(10) + 400 &= 5(100) - 900 + 400\
&= 500 - 900 + 400\
&= 0
\end{align*}
$$
Since the result is $ 0 $, $ x = 10 $ is a solution.

## Step3: Check $ x = -8 $
Substitute $ x = -8 $ into the equation $ 5x^2 - 90x + 400 = 0 $.
$$
\begin{align*}
5(-8)^2 - 90(-8) + 400 &= 5(64) + 720 + 400\
&= 320 + 720 + 400\
&= 1440
\end{align*}
$$
Since $ 1440
eq 0 $, $ x = -8 $ is not a solution.

# Answer:
- $ x = 0 $: Not a solution (substitution gives $ 400
eq 0 $).
- $ x = 10 $: Solution (substitution gives $ 0 $).
- $ x = -8 $: Not a solution (substitution gives $ 1440
eq 0 $).

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QUESTION IMAGE

Question

Explanation:

Step1: Check \( x = 0 \)

Step2: Check \( x = 10 \)

Step3: Check \( x = -8 \)

Answer: