Can somebody complete the square in both x and y for the following equation:
3x²+18x+2y²-8y=32
I still need to figure out how to do this, so if you could show me how it's done, I'd appreciate it. :)
3x²+18x+2y²-8y=32
I still need to figure out how to do this, so if you could show me how it's done, I'd appreciate it. :)
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Oh I'm doing this in school right now too.
First you take out the 2 and three from x and y respectively, so
3(x^2 +6x) + 2(y^2 -4x) = 32.
Then to complete square you divide the second number in the parentheses by two and then square it. So the 6 leads to a 9 and the -4 leads to a 4. Like so.
3(x^2 +6x +9) + 2(y^2 -4x +4) = 32
However you also have to add these newly added numbers to the right side of the equation. But you have to multiply the coefficient by the 9 and 4. So add a 3(9) and a 2(4) to the 32. You get 67.
So 3(x +3)^2 + 2(y-2)^2 = 67
Then divide both sides by 67 so the number on the right becomes a 1. If it had been a 66 your value under the 3(x +3)^2 would be 22 and the value under (y-3)^2 would be a 33. However since it doesn't add up right I suspect you might not have typed it in correctly. But that's what it would look like in standard form.
First you take out the 2 and three from x and y respectively, so
3(x^2 +6x) + 2(y^2 -4x) = 32.
Then to complete square you divide the second number in the parentheses by two and then square it. So the 6 leads to a 9 and the -4 leads to a 4. Like so.
3(x^2 +6x +9) + 2(y^2 -4x +4) = 32
However you also have to add these newly added numbers to the right side of the equation. But you have to multiply the coefficient by the 9 and 4. So add a 3(9) and a 2(4) to the 32. You get 67.
So 3(x +3)^2 + 2(y-2)^2 = 67
Then divide both sides by 67 so the number on the right becomes a 1. If it had been a 66 your value under the 3(x +3)^2 would be 22 and the value under (y-3)^2 would be a 33. However since it doesn't add up right I suspect you might not have typed it in correctly. But that's what it would look like in standard form.
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3x²+18x+2y²-8y=32
{3(x^2+6x +9)-27} +{2(y^2-4y+4)-8} =32
3(x+3)^2+2(y+2)^2 =32+27+8
3(x+3)^2 +2(y+2)^2 =67
Divide by 67
{(x+3)^2/(67/3)} +{y+2)^2/(67/2)}^2 = 1 ...............Ans
{3(x^2+6x +9)-27} +{2(y^2-4y+4)-8} =32
3(x+3)^2+2(y+2)^2 =32+27+8
3(x+3)^2 +2(y+2)^2 =67
Divide by 67
{(x+3)^2/(67/3)} +{y+2)^2/(67/2)}^2 = 1 ...............Ans
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3[x^2+6x+9]+2[y^2-4y+4] = 32+27+8
3(x+3)^2+2(y-2)^2 = 67
3(x+3)^2+2(y-2)^2 = 67