Solve x2 − 8x + 5 = 0 using the completing-the-square method.
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OpenStudy (ofmiceparade):
@hartnn Hi
hartnn (hartnn):
hello :)
what have you tried?
OpenStudy (ofmiceparade):
i got x and square root of 11
OpenStudy (ofmiceparade):
I don't think i got it right
hartnn (hartnn):
ok, lets go step by step,
x^2 -8x = -5
what did you add to both sides of this equation?
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OpenStudy (ofmiceparade):
16
hartnn (hartnn):
good!
so left side became the perfect square of (x-4)
\((x-4)^2 = 16-5
\\ (x-4)^2=11\)
now take square root on both sides
OpenStudy (ofmiceparade):
?
OpenStudy (ofmiceparade):
11 can't be square rooted
OpenStudy (ofmiceparade):
and idk about x-4
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hartnn (hartnn):
right,
keep it as \(\Large \sqrt{11}\)
OpenStudy (ofmiceparade):
okay
hartnn (hartnn):
\(\sqrt{(x-4)^2} = x-4\)
OpenStudy (ofmiceparade):
okay so x-4 = square root of 11
hartnn (hartnn):
\(\Large x-4 =\pm \sqrt {11}\)
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hartnn (hartnn):
because there would be a negative root too
OpenStudy (ofmiceparade):
okay but my answer choices are
x = negative four plus or minus the square root of eleven
x = four plus or minus the square root of eleven
x = negative four plus or minus the square root of five
x = four plus or minus the square root of five
OpenStudy (ofmiceparade):
That isn't "x ="
hartnn (hartnn):
add 4 on both sides
OpenStudy (ofmiceparade):
OHHHH
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hartnn (hartnn):
\(\Large x = 4\pm \sqrt{11}\)
OpenStudy (ofmiceparade):
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