Asked by Kimberly Williams
Solve and check each radical equation. square root y+10=y-2
Answers
Answered by
Jai
sqrt(y + 10) = y - 2
First thing to do is to square both sides of the equation, then solve for x:
y + 10 = (y - 2)^2
y + 10 = y^2 - 4y + 4
0 = y^2 - 5y - 6
0 = (y - 6)(y + 1)
y = 6
y = -1
Now to check, substitute the x values back to original equation.
y = -1:
sqrt(-1 + 10) = -1 - 2
sqrt(9) = -3
3 = -3
Thus x is NOT equal to -1. This is an extraneous root.
y = 6:
sqrt(6 + 10) = 6 - 2
sqrt(16) = 4
4 = 4
Thus x is indeed equal to 6.
Hope this helps~ :3
First thing to do is to square both sides of the equation, then solve for x:
y + 10 = (y - 2)^2
y + 10 = y^2 - 4y + 4
0 = y^2 - 5y - 6
0 = (y - 6)(y + 1)
y = 6
y = -1
Now to check, substitute the x values back to original equation.
y = -1:
sqrt(-1 + 10) = -1 - 2
sqrt(9) = -3
3 = -3
Thus x is NOT equal to -1. This is an extraneous root.
y = 6:
sqrt(6 + 10) = 6 - 2
sqrt(16) = 4
4 = 4
Thus x is indeed equal to 6.
Hope this helps~ :3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.