3x2 -6x - 4 =0 how to complete the square?

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3x2 -6x - 4 =0 how to complete the square?

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Answer 1 · 2018-02-16T12:29:55

See below...

Explanation:

We have the quadratic $3 x^{2} - 6 x - 4 = 0$ $3x^2-6x-4=0$

First of all, we take out a factor of 3. Don't take it out from the constant however as it can lead to some unnecessary fraction work.

$3 x^{2} - 6 x - 4 \Rightarrow 3 [x^{2} - 2 x] - 4$ $3x^2-6x-4 => 3[x^2-2x]-4$

Now we write out our initial bracket. To do this we have

${(x + \frac{b}{2})}^{2}$ $(x+b/2)^2$ $\Rightarrow$ $=>$ in this case $b$ $b$ is $- 2$ $-2$ . Note that we don't include an $x$ $x$ after the $b$ $b$ ...
Once we have our initial bracket, we subtract the square of $\frac{b}{2}$ $b/2$

$therefore 3[x^2-2x]-4 => 3[(x-1)^2 -1]-4$

Now we must remove the square brackets by multiplying whats in it by the factor on the outside, in this case $3$ .

$therefore$ we get $3(x-1)^2 -3-4 = 3(x-1)^2 -7$

Final answer $3(x-1)^2 - 7 =0$

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3x2 -6x - 4 =0 how to complete the square?

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Explanation:

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