SOLUTION: PROBLEM: SOLVE THE EQUATION BY COMPLETING THE SQUARE AND APPLYING THE SQUARE ROOT PROPERTY. 4y^2-12y+13=0 THIS PROBELM IS VERY DIFFICULT FOR ME TO UNDERSTAND I KNOW THIS MUCH SO

Algebra ->  Matrices-and-determiminant -> SOLUTION: PROBLEM: SOLVE THE EQUATION BY COMPLETING THE SQUARE AND APPLYING THE SQUARE ROOT PROPERTY. 4y^2-12y+13=0 THIS PROBELM IS VERY DIFFICULT FOR ME TO UNDERSTAND I KNOW THIS MUCH SO      Log On


   

Question 138827: PROBLEM:
SOLVE THE EQUATION BY COMPLETING THE SQUARE AND APPLYING THE SQUARE ROOT PROPERTY.
4y^2-12y+13=0
THIS PROBELM IS VERY DIFFICULT FOR ME TO UNDERSTAND I KNOW THIS MUCH SO FAR BUT I GET STUCK CAN YOU PLEASE HELP!!
4y^2-12y=-13
4y^2-12y+(-12/2)^2=-13+(-12-2)^2
this is as far as i can get please help
Found 2 solutions by vleith, stanbon:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
%28ay-b%29%5E2+ = a%5E2y%5E2+-+2aby+%2B+b%5E2
You are given 4y%5E2-12y+=+-13
In this equation, a is 2 (since 4 = 2^2)
And -12+=+-2ab = -12+=+-4b = 3+=+b
4y%5E2-12y+%2B9+=+-13+%2B+9
%282y-3%29%5E2+=+-4
2y-3+=+%2B-2i
y+=+%283%2B2i%29%2F2 and
y+=+%283-2i%29%2F2


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
4y^2-12y+13=0
4y^2 - 12y = -13
4(y^2 - 3y + ?) = -13 + 4*?
4(y^2-3y+(3/2)^2) = -13 + 4*(3/2)^2
4(y - (3/2))^2 = -13 +9
Divide both sides by 4 to get:
(y-(3/2))^2 = -1
Take the square root of both sides to get:
y-(3/2) = +/- i

y = [(3/2) + i] or y = [(3/2) - i]

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Cheers,
Stan H.