Asked by ki
SOLVE THE FOLLOWING EQAUTION BY COMPLETING THE SQUARE:
3. xsquared - 8x - 10 = 0
xsqaured -8x=10
xsquared-8x+(8/2)sqaured=10+(8/2)sqaured
xsquared-8x+16=26
Answer: xsqaured-8x+16=26
WHAT DID I DO WRONG?
3. xsquared - 8x - 10 = 0
xsqaured -8x=10
xsquared-8x+(8/2)sqaured=10+(8/2)sqaured
xsquared-8x+16=26
Answer: xsqaured-8x+16=26
WHAT DID I DO WRONG?
Answers
Answered by
Reiny
I will do it without "words"
x^2 - 8x = 10
x^2 - 8x + 16 = 10+16
(x-4)^2 = 26
x-4 = ± √26
x = 4 ± √26
you stopped and didn't finish it
Where was your "completed square" ?
x^2 - 8x = 10
x^2 - 8x + 16 = 10+16
(x-4)^2 = 26
x-4 = ± √26
x = 4 ± √26
you stopped and didn't finish it
Where was your "completed square" ?
Answered by
ki
i don't know what that means
Answered by
ki
is that the answer
Answered by
Reiny
Are you taking this topic in a regular class-room setting?
If you are studying the topic of "completing the square", which was part of the instructions to your problem, you MUST know what this means.
If you are studying the topic of "completing the square", which was part of the instructions to your problem, you MUST know what this means.
Answered by
Reiny
yes, that is the answer.
you could say
x = 10+√26 OR x = 10 - √26
or you could convert to decimals with your calculator.
Leaving it in square root form is considered exact, while as soon as you change square roots to decimals you are dealing with approximations.
you could say
x = 10+√26 OR x = 10 - √26
or you could convert to decimals with your calculator.
Leaving it in square root form is considered exact, while as soon as you change square roots to decimals you are dealing with approximations.
Answered by
Reiny
that should have said:
you could say
x = 4+√26 OR x = 4 - √26
you could say
x = 4+√26 OR x = 4 - √26
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