Asked by Student
Will you please check my work?
I'm supposed to "Use square roots to solve the following equations; round to the nearest hundredth, if necessary."
1. 5x^2 - 256 = -651
My Work:
5x^2 - 526 + 526 = -651 + 526
(sq root of) 5x^2 = (sq root of) -125 = 5x/5 = 11.18033989/5 = 2.24
2. 3z^2 - 22 = 56
My Work:
3z^2 - 22 + 22 = 56 + 22 =
(sq root of)3z^2 = (sq root of)78 = 3z/3 = 8.831760866 = 2.94
Did I do these correctly? Also, (as dumb as it seems) I sometimes have trouble rounding, so can you make sure I did that correctly, as well?
Thanks so much!
I'm supposed to "Use square roots to solve the following equations; round to the nearest hundredth, if necessary."
1. 5x^2 - 256 = -651
My Work:
5x^2 - 526 + 526 = -651 + 526
(sq root of) 5x^2 = (sq root of) -125 = 5x/5 = 11.18033989/5 = 2.24
2. 3z^2 - 22 = 56
My Work:
3z^2 - 22 + 22 = 56 + 22 =
(sq root of)3z^2 = (sq root of)78 = 3z/3 = 8.831760866 = 2.94
Did I do these correctly? Also, (as dumb as it seems) I sometimes have trouble rounding, so can you make sure I did that correctly, as well?
Thanks so much!
Answers
Answered by
Student
I wrote the first problem incorrectly. It's supposed to be 5x^2 - 526 = -651
I apologize.
I apologize.
Answered by
Reiny
5x^2 - 526 = -651
5x^2 = -125
x^2 = -25
Unless you have another typo, there is no real number solution, since we cannot take the square root of a negative.
If you equation was supposed to be
5x^2 - 526x= -651
5x^2 - 526x + 651 = 0
x = (526 ± √263656)/10
= appr 103.95 or appr 1.25
2nd:
3z^2- 22 = 56
3z^2 = 78
z^2 = 26
z ± √26 = ± 5.09901..= appr ±5.10
5x^2 = -125
x^2 = -25
Unless you have another typo, there is no real number solution, since we cannot take the square root of a negative.
If you equation was supposed to be
5x^2 - 526x= -651
5x^2 - 526x + 651 = 0
x = (526 ± √263656)/10
= appr 103.95 or appr 1.25
2nd:
3z^2- 22 = 56
3z^2 = 78
z^2 = 26
z ± √26 = ± 5.09901..= appr ±5.10
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