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Mathematics 20 Online
OpenStudy (anonymous):

x =√(24 − 5x)

OpenStudy (anonymous):

This one should also give you 2 answers, I don't know what to tell you.

OpenStudy (anonymous):

Ugh, that's what I thought. Maybe as a decimal it'll work.

jimthompson5910 (jim_thompson5910):

x = sqrt(24 - 5x) x^2 = 24 - 5x x^2 + 5x - 24 = 0 Now use the quadratic formula to solve x = (-b+-sqrt(b^2-4ac))/(2a) x = (-(5)+-sqrt((5)^2-4(1)(-24)))/(2(1)) x = (-5+-sqrt(25-(-96)))/(2) x = (-5+-sqrt(121))/2 x = (-5+sqrt(121))/2 or x = (-5-sqrt(121))/2 x = (-5+11)/2 or x = (-5-11)/2 x = 6/2 or x = -16/2 x = 3 or x = -8 So the possible solutions are x = 3 or x = -8 However, x = -8 is a extraneous solution because the range of the square root function is always nonnegative. So the only solution is x = 3

OpenStudy (anonymous):

Okay awesome, I'm starting to understand these things. Thanks a bunch.

OpenStudy (anonymous):

After you perform the operations under the sq root, 24 -5(-8), the answer is still real. -8 is also a solution.

OpenStudy (anonymous):

Well the answer was 3 regardless. Not sure why anymore though.

jimthompson5910 (jim_thompson5910):

No -8 is not a solution, plug it back into the original equation and you'll see that x = sqrt(24 - 5x) -8 = sqrt(24 - 5(-8)) -8 = sqrt(24 + 40) -8 = sqrt(64) -8 = 8 which is false So again, x = -8 is NOT a solution

OpenStudy (anonymous):

oops my mistake.

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