# 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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