Of course, let's solve the given equation step-by-step:
Given equation:
[tex]\[ x - 3 = \sqrt{x + 9} \][/tex]
1. Isolate the square root:
The equation is already in a form where one side is free of radicals.
2. Square both sides of the equation:
To eliminate the square root, square both sides:
[tex]\[ (x - 3)^2 = (\sqrt{x + 9})^2 \][/tex]
Simplifying this gives:
[tex]\[ x^2 - 6x + 9 = x + 9 \][/tex]
3. Move all terms to one side to form a quadratic equation:
Subtract [tex]\( x + 9 \)[/tex] from both sides:
[tex]\[ x^2 - 6x + 9 - x - 9 = 0 \][/tex]
Simplify:
[tex]\[ x^2 - 7x = 0 \][/tex]
4. Factor the quadratic equation:
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(x - 7) = 0 \][/tex]
5. Solve for [tex]\( x \)[/tex]:
Set each factor to zero:
[tex]\[ x = 0 \][/tex]
[tex]\[ x - 7 = 0 \][/tex]
[tex]\[ x = 7 \][/tex]
6. Check for extraneous solutions:
Substitute each solution back into the original equation to ensure they satisfy it.
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ 0 - 3 = \sqrt{0 + 9} \][/tex]
[tex]\[ -3 = 3 \][/tex]
This is false, so [tex]\( x = 0 \)[/tex] is an extraneous solution.
- For [tex]\( x = 7 \)[/tex]:
[tex]\[ 7 - 3 = \sqrt{7 + 9} \][/tex]
[tex]\[ 4 = \sqrt{16} \][/tex]
[tex]\[ 4 = 4 \][/tex]
This is true, so [tex]\( x = 7 \)[/tex] is a valid solution.
Thus, the equation [tex]\( x - 3 = \sqrt{x + 9} \)[/tex] has one valid solution:
[tex]\[
\boxed{7}
\][/tex]
