Question

hello guyz i need help

4a^2 + 4a -3 = 0

2z^2 - z - 6 = 0

solve by completing the square

Answer 1

phew, It's been a long time since I didin't complete a square, but let's analyze it.
by definition:
\[a ^{2}+2ab+b ^{2}=(a+b)^{2}\]
But we'll center in this little chunk:
\[(a+b)^{2}=0\]
So by hankelian property we know:
\[a+b=0 =>a=-b\]
So, not so clear, but I'll say, this little fellow must have roots, that respect the general formula. Meaning that, I could use the expanded form to find those roots and these should be double.
But, not so long ago, there was a little trick for these, I'll use the first one to show it to you.
\[4a ^{2}+4a-3=0\]
I'll move the 3 to the other side of the equality, and then divide both sides by 4:
\[a ^{2}+a=\frac{ 3 }{ 4 }\]
Now, I shall take the simple exponent a, and divide it by 2, and add that amount to both sides:
\[a ^{2}+a+\frac{ 1 }{ 4 }=\frac{ 3 }{ 4 }+\frac{ 1 }{ 4 }\]
Now I have a completed square on one side, and a fractional addition on the other, let's solve them:
\[(a+\frac{ 1 }{ 2 })^{2}=1\]
Now, let's take square root, and knowing that square root of 1, is 1:
\[a+\frac{ 1 }{ 2 }=1\]
\[a=1-\frac{ 1 }{ 2 }\]
\[a=\frac{ 1 }{ 2 }\]

Answer 2

sorry, forgot the:
\[a=\pm \frac{ 1 }{ 2 }\]

Answer 3

hey @owlcoffee im stuck here can you help me

\[z ^{2}-\frac{ z }{ 2 }=3\]

Answer 4

@owlcoffee thanks by the way on explaining how to complete the square

Answer 5

@Abdulhameed any ideas?

Answer 6

\[x ^{2}-\frac{ x }{ 2}=3\]
\[\frac{ 2x ^{2} -x}{2 }=3\]
\[6=2x ^{2}-x\]
\[2x ^{2}-x-6=0\]
\[(x-2)(2x+3)=0\]
either
x-2=0 then x=2
or 2x+3=0 then 2x=-3 , x=-3/2

Answer 7

@denizen

Answer 8

\[x ^{2}-\frac{ x }{ 2 }=3\]
Same method, since x^2 already has a coefficent of 1, we'll divide the x term by 2, square it, and add it on both sides:
\[x ^{2}-\frac{ 1 }{ 2 }x=3\]
\[x ^{2}-\frac{ 1 }{ 2 }x+\frac{ 1 }{ 16 }=3+\frac{ 1 }{ 16 }\]
Completing the squares and doing the fractionary addition on the right side:
\[(x-\frac{ 1 }{ 4 })^{2}=\frac{ 49 }{ 16 }\]
Now let's solve for x:
\[x-\frac{ 1 }{ 4 }=(+/-)\sqrt{\frac{ 49 }{ 16 }}\]
I will not prove it, (because I'm sleepy, haven't slept in like... 2 days) but the square root of a fraction is equal to the square root of numerator and denominator.
Square root of 49 is 7, and square root of 16 is 4.
so:
\[x=\frac{ 1 }{ 4 }\pm \frac{ 7 }{ 4 }\]
or, for more confort:
\[x=\frac{ 1\pm7 }{ 4 }\]
Please, recheck what I did, because I have a lack of sleep and when I'm sleepy I make huge mistakes like 4/2=1/2 (actually did that on a test, I'm such a fail ): )