Question

Consider the equation:
\[24=x^2-4x+3 \]
1) Rewrite the equation by completing the square.
Your equation should look like
\[(x+c)^2=d\] or
\[(x-c)^2=d\].
2) What are the solutions to the equation?
Choose 1 answer:
Choose 1 answer:
(Choice A)
\[x=2 \pm 5\]
A
\[x=2 \pm 5\]
(Choice B)
\[x=-2 \pm 5\]
B
\[x=-2 \pm 5\]
(Choice C)
\[x=2 \pm \sqrt{5}\]
C
\[x=2 \pm \sqrt{5}\]
(Choice D)
\[x=-2 \pm \sqrt{5}\]
D
\[x=-2 \pm \sqrt{5}\]

Answer 1

To solve the equation \(24 = x^2 - 4x + 3\), we start by moving all terms to one side:

\[
x^2 - 4x + 3 - 24 = 0
\]

Simplifying gives:

\[
x^2 - 4x - 21 = 0
\]

Next, we will complete the square for the quadratic expression \(x^2 - 4x\). The process involves taking the coefficient of \(x\), which is \(-4\), halving it to get \(-2\), and then squaring it to find \(4\). Let's rewrite the equation:

1. Rewrite:

\[
x^2 - 4x = (x - 2)^2 - 4
\]

Now we can substitute this back into our equation:

\[
(x - 2)^2 - 4 - 21 = 0
\]

Combining the constants gives:

\[
(x - 2)^2 - 25 = 0
\]

Next, we can reorganize the equation to be in the form \((x - c)^2 = d\):

\[
(x - 2)^2 = 25
\]

2. Now, we solve for \(x\):

Taking the square root of both sides:

\[
x - 2 = \pm 5
\]

This gives us two equations to solve:

\[
x - 2 = 5 \quad \text{or} \quad x - 2 = -5
\]

Solving these:

1. For \(x - 2 = 5\):

\[
x = 5 + 2 = 7
\]

2. For \(x - 2 = -5\):

\[
x = -5 + 2 = -3
\]

Thus, the solutions are:

\[
x = 2 \pm 5
\]

From the options provided, the correct choice is:

**(Choice A) \(x=2 \pm 5\)**.