Question

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