Asked by xdcfv
solve the equation by completing the square. Round to the nearest tenth. x^2+2x=8
Answers
Answered by
Bot
To complete the square, we need to add and subtract the square of half the coefficient of x, which is (2/2)^2 = 1:
x^2 + 2x + 1 - 1 = 8
Now we can simplify:
(x+1)^2 - 1 = 8
Add 1 to both sides:
(x+1)^2 = 9
Take the square root of both sides:
x+1 = ±√9
x+1 = ±3
Subtract 1 from both sides:
x = -1±3
We get two solutions: x = 2 and x = -4
Rounding to the nearest tenth is not necessary since the solutions are integers. Therefore, the solutions are x = 2 and x = -4.
x^2 + 2x + 1 - 1 = 8
Now we can simplify:
(x+1)^2 - 1 = 8
Add 1 to both sides:
(x+1)^2 = 9
Take the square root of both sides:
x+1 = ±√9
x+1 = ±3
Subtract 1 from both sides:
x = -1±3
We get two solutions: x = 2 and x = -4
Rounding to the nearest tenth is not necessary since the solutions are integers. Therefore, the solutions are x = 2 and x = -4.
Answered by
Bot
Yes, that's correct! The solutions are x = 2 and x = -4.
Answered by
Bot
You're welcome! Don't hesitate to ask if you have any more questions.
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