Determine whether the values 3 and 4 are solutions to the quadratic equation

Question

$x^{2}-7x+12=0$ and then select the correct answer from the following options.
Option \#1: Only $x=3$ is a solution.
Option \#2: Only $x=4$ is a solution.
Option \#3: Both $x=3$ and $x=4$ are solutions.
Option \#4: Neither $x=3$ nor $x=4$ is a solution.

Answer 1

To determine whether the values 3 and 4 are solutions to the quadratic equation \( x^{2}-7x+12=0 \), we can substitute these values into the equation and see if either satisfies it.

First, let’s check \( x = 3 \):

\[ x^2 - 7x + 12 = 0 \] Substituting \( x = 3 \): \[ 3^2 - 7(3) + 12 = 0 \] This simplifies to: \[ 9 - 21 + 12 = 0 \] Calculating further: \[ 9 - 21 + 12 = 0 \ 0 = 0 \] This shows that \( x = 3 \) is indeed a solution.

Next, let’s check \( x = 4 \):

Substituting \( x = 4 \): \[ 4^2 - 7(4) + 12 = 0 \] This simplifies to: \[ 16 - 28 + 12 = 0 \] Calculating further: \[ 16 - 28 + 12 = 0 \ 0 = 0 \] This shows that \( x = 4 \) is also a solution.

Since both values \( x = 3 \) and \( x = 4 \) satisfy the equation, the correct answer is:

Option #3: Both \( x=3 \) and \( x=4 \) are solutions.