The extraneous solution of the given radical equation is x = 3
Here we have the radical equation:
[tex]\sqrt{45 - 3x} = x - 9[/tex]
We want to get the extraneous solution, to get it, we first need to square both sides of the equation, so we get:
[tex](\sqrt{45 - 3x} )^2 = (x - 9)^2\\\\45 - 3x = x^2 - 18x + 81[/tex]
Now we got a quadratic equation, we can rewrite it as:
[tex]x^2 - 18x + 81 + 3x - 45 = 0\\\\x^2 - 15x + 36 = 0[/tex]
To solve this we use Bhaskara's formula, we will get:
[tex]x = \frac{15 \pm \sqrt{(-15)^2 - 4*1*36} }{2*1} \\\\x = \frac{15 \pm 9}{2}[/tex]
Then the two solutions are:
x = (15 + 9)/2 = 12
x = (15 - 9)/2 = 3
Let's evaluate both solutions in our original equation and let's see which is the one that does not work:
for x = 12 we have:
[tex]\sqrt{45 - 3*12} = 12 - 9\\3 = 3[/tex]
This is true.
For x = 3 we have:
[tex]\sqrt{45 - 3*3} = 3 - 9\\\\6 = -6[/tex]
This is false, so the extraneous solution is x = 3.
If you want to learn more about extraneous solutions, you can read:
https://brainly.com/question/2959656
