Question

Which of the following is an extraneous solution of (45 minus 3 x) Superscript one-half Baseline = x minus 9?

Answer 1

The extraneous solution of the given radical equation is x = 3

Finding the extraneous solution:

Here we have the radical equation:

$\sqrt{45 - 3x} = x - 9$

We want to get the extraneous solution, to get it, we first need to square both sides of the equation, so we get:

$(\sqrt{45 - 3x} )^2 = (x - 9)^2\\\\45 - 3x = x^2 - 18x + 81$

Now we got a quadratic equation, we can rewrite it as:

$x^2 - 18x + 81 + 3x - 45 = 0\\\\x^2 - 15x + 36 = 0$

To solve this we use Bhaskara's formula, we will get:

$x = \frac{15 \pm \sqrt{(-15)^2 - 4*1*36} }{2*1} \\\\x = \frac{15 \pm 9}{2}$

Then the two solutions are:

x = (15 + 9)/2 = 12

x = (15 - 9)/2 = 3

Which one is the extraneous solution?

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:

$\sqrt{45 - 3*12} = 12 - 9\\3 = 3$

This is true.

For x = 3 we have:

$\sqrt{45 - 3*3} = 3 - 9\\\\6 = -6$

This is false, so the extraneous solution is x = 3.

If you want to learn more about extraneous solutions, you can read:

brainly.com/question/2959656