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Complete the square -3x^2+12x-7

Sagot :

Answer:

[tex]-3(x-2)^2+5[/tex]

Step-by-step explanation:

First we can factor a -3 from the [tex]x^2[/tex] term and the [tex]x[/tex] term to get [tex]-3(x^2-4x)-7[/tex].

Then we want the stuff in the parentheses to have the form of [tex](x+b)^2[/tex], or equivalently, [tex](x^2+2b+b^2)[/tex] . So we can let [tex]2b = -4[/tex]. By solving it, we get [tex]b = -4/2 = -2[/tex]. Then our [tex]b^2[/tex] term should be [tex]b^2 = (-2)^2 = 4[/tex].

In order to make our [tex]b^2[/tex] term appear in the parentheses, we need to add and subtract our [tex]b^2[/tex] term, so we get [tex]-3(x^2 - 4x + 4 - 4) -7[/tex].

What we to keep inside our parentheses is [tex](x^2 - 4x + 4)[/tex] , so we can factor the [tex]-4[/tex] out of parentheses to get [tex]-3(x^2-4x+4-4)-7 = -3(x^2-4x+4)+(-3)(-4) - 7 = -3(x^2 - 4x + 4) + 12 - 7 = -3(x^2 - 4x + 4) + 5[/tex]

Finally, plugging  [tex]b = -2[/tex]  that we computed earlier into the equation [tex](x^2+2b+b^2) = (x+b)^2[/tex], we get  [tex](x^2 - 4x + 4) = (x-2)^2[/tex].

So we have [tex]-3(x^2-4x+4)+5 = -3(x-2)^2+5[/tex].

In summary, the procedure is

[tex]-3x^2+12x-7 \\= &-3(x^2-4x) -7 \\= &-3(x^2-4x+4-4)-7 \\= -3(x^2-4x+4) + (-3)(-4)-7\\=-3(x^2-4x+4)+12-7\\= -3(x^2-4x+4)+5 \\= -3(x-2)^2+5[/tex]