Find x: √(x^2 - 9) = x + 1?
I did this equation for homework but I'm not sure if it's right.
So first I squared the left side and the right side to cancel out the square root on the left.
So I get:
x^2 - 9 = (x + 1)^2 then
x^2 - 9 = x^2 + 2x + 1
Then I subtracted x^2 and added 9 over to the right so it equals to 0.
When I subtracted x^2 over, it cancelled out the x^2 so now I have:
2x + 10 = 0
Then I subtracted 10 over and then divided 2 so I get:
x = -5
Is my work right?? Thanks for the help!!!!
Update:Thanks everyone! and @Iggy Rocko righttt I have to check for -5, i forgot! thanks! so therefore, there is no solution :)
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Verified answer
x^2 - 9 = x^2 + 2x + 1
-9 = 2x + 1
-10 = 2x
x = -5
The problem is substituting -5 back into the original equation yields a false result.
√((-5)^2 - 9) = -5 + 1
√(25 - 9) = -4
√16 = -4
4 = -4 which is false
√ always yields a non-negative result. Therefore, there is no solution.
Technically, you should have ±(x + 1)^2 on the right side.
Then you have:
x^2 - 9 = (x + 1)^2...and....x^2 - 9 = -(x + 1)^2
=> x^2 - 9 = x^2 + 2x + 1 ----- you got this one
and
=> x^2 - 9 = -(x^2 + 2x + 1) = -x^2 - 2x - 1
=> 2x^2 + 2x - 8 = 0
=> x = (1/2)(-1 ± â17)
Neither answer gives a valid result. You can see this by looking at the solution "-5"...under the radical sign you get "-5 - 9" which is a negative, giving an imaginary number.
So, this problem has no solution.
These are tricky....you have to be very careful working with radical-problems.
x^2-9 = x^2+2x+1
2x = -10
x = -5
Looks good.
Yes, you are right,
good job