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Answer to Question #22482 in Trigonometry for Kristen Woods
Question #22482
Solve the given equations by completing the square.
6a. x^2+2x+5=0
6b. 2x^2-7x+6=0
6a. x^2+2x+5=0
6b. 2x^2-7x+6=0
Expert's answer
6a. x^2+2x+5=0
x^2+2x+1=-4
(x+1)^2=-4
x+1=+-2i
x1=-1+2i, x2=-1-2i
6b. 2x^2-7x+6=0
x^2-7/2x+3=0
x^2-2*(7/4)*x+(7/4)^2-(7/4)^2+3=0
(x-7/4)^2=(7/4)^2-3
(x-7/4)^2=1/16
x-7/4=+-1/4
x1=7/4+1/4, x2=7/4-1/4
x1=8/4=2, x2=6/4=3/2
x^2+2x+1=-4
(x+1)^2=-4
x+1=+-2i
x1=-1+2i, x2=-1-2i
6b. 2x^2-7x+6=0
x^2-7/2x+3=0
x^2-2*(7/4)*x+(7/4)^2-(7/4)^2+3=0
(x-7/4)^2=(7/4)^2-3
(x-7/4)^2=1/16
x-7/4=+-1/4
x1=7/4+1/4, x2=7/4-1/4
x1=8/4=2, x2=6/4=3/2
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