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# Precalculus

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Can someone help complete the square?

4m^2 - 12m + 13 = 0

thank you!

Guest Sep 2, 2017
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#1
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Can someone help complete the square?

4m^2 - 12m + 13 = 0

Solve for m:
4 m^2 - 12 m + 13 = 0

Divide both sides by 4:
m^2 - 3 m + 13/4 = 0

Subtract 13/4 from both sides:
m^2 - 3 m = -13/4

Add 9/4 to both sides:
m^2 - 3 m + 9/4 = -1

Write the left hand side as a square:
(m - 3/2)^2 = -1

Take the square root of both sides:
m - 3/2 = i or m - 3/2 = -i

Add 3/2 to both sides:
m = 3/2 + i or m - 3/2 = -i

Add 3/2 to both sides:
m = 3/2 + i          or             m = 3/2 - i

Guest Sep 2, 2017
edited by Melody  Sep 2, 2017
#2
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\(\displaystyle 4m^{2}-12m+13\\=4(m^{2}-3m)+13\\=4(m^{2}-3m+9/4-9/4)+13\\=4\{(m-3/2)^{2}-9/4\}+13 \\=4(m-3/2)^{2}-9+13\\=4(m-3/2)^{2}+4.\)

Guest Sep 3, 2017

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