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how do you complete the square in the equation 4y^2-2x+8y+1=0?
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Okay. I think I’ve got it. First off, move all your y’s to one side of the equation, and all your x’s to the other. 4y^2 + 8y = 2x – 1 Next step, you have to make 1 the coefficient of your y^2! So divide everything by 4. y^2 + 2y = 2x/4 – ¼ Now, take the coefficient of your y^2 (which we made 1), and multiply it by the coefficient of y. That gives you 1. Insert this number on both sides of the equation! y^2 + 2y + 1 = 2x/4 – ¼ + 1 Now you can make the left-hand side of the equation (y+1)^2! All you have to do now is rearrange stuff in the formula, isolating the x. 2 (y + 1)^2 – 6/4 = x I hope I haven’t made any mistakes!
thanx a million!
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