Which of the following is an extraneous solution of (45 minus 3 x) Superscript one-half Baseline = x minus 9? x = –12 x = –3 x = 3 x = 12

x = 3

Step-by-step explanation:

I guess that the equation is:

(45 - 3x)^1/2 = x - 9

so let's solve it for x.

first, we can square both sides:

(45 - 3x) = (x - 9)^2 = x^2 - 18x + 81

now we can write this as a quadratic equation:

x^2 - 18x + 81 - 45 + 3x = 0

x^2 -15x + 36 = 0

now we can use the Bhaskara's equation to find the solutions for that equation:

where for a equation a*x^2 + b*x + c = 0

the solutions are:

here a = 1, b = -15 and c = 36

then the solutions are:

x = (15 + 9)/2 = 24/2 = 12

x = (15 - 9)/2 = 6/2 = 3

where 12 is the solution for the positive (45 - 3x)^1/2 and x = 3 is the extraneous solution (because it works for the negative  (45 - 3x)^1/2)

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step-by-step explanation:

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