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How to “Re-write completing the square”: $x^2+x+1$

$b^2-4b+4=16$ can someone help me solve this equation. All the steps please.


marked as duplicate by Qiaochu Yuan May 1 '11 at 20:35

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  • 1
    $\begingroup$ Since you say this is a homework problem, you really shouldn't ask for all the steps. What have you tried? Where are you getting stuck? Do you know what "completing the square" means? What difficulty are you having applying that method? The more information you give about your problem, the more helpful people will want to be, and, quite possibly, the more you will learn. $\endgroup$ – Matthew Conroy Apr 1 '11 at 0:04
  • $\begingroup$ See also my answer here for a particular step-by-step method of completing the square on a quadratic expression. $\endgroup$ – Isaac May 1 '11 at 20:41

if $ax^2+bx^2+c=0$ then $$ 0=a\Big(x^2+\frac{b}{a}x\Big)+c =a\Big(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\Big)+c =a\Big(x+\frac{b}{2a}\Big)^2+c-\frac{b^2}{4a} $$ hence $$ \Big(x+\frac{b}{2a}\Big)^2=\frac{b^2-4ac}{4a^2} $$ and $$ x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$


The idea of completing the square is to get a simple expression such as

(b-something-you-know)^2 = something-you-also-know

Then you can just take the square-root on both sides, on you have your solution. In the example you give, you do not really have to do a lot of work. Usually, one looks at the terms with the variables, in your case


You should see that this is part of what you get if you have the expression

(b-2)^2 (the second binomial formula)

Then you assume that this formula actually appears on the left-hand side of your original equation. Of course, you have to make sure that you do not change the equation, so you multiply it out and see whether there is some difference to be accounted for.

(b-2)^2 = b^2 - 2*b*2 + 2^2 = b^2 - 4b + 4

In fact, there is no difference, so you can just turn your expression into

(b-2)^2 = 16

which, in turn, reduces to

(b1-2) = 4


(b2-2) = -4 (remember that quadratic equations can have two solutions)

and you get b1 = 6 and b2=-2

You would have needed to actually complete the square if your original example had read, say,

b^2 - 4b - 5 = 16

again, you would assume that on the left-hand side, there is (b-2)^2 (from looking just at b^2-4b).

Now, however, there is a difference, namely the difference between

1) b^2 - 4b - 5 (the original left-hand side)


2) b^2-4b+4 (the left-hand side you want to have, i.e. (b-2)^2)

But there is an easy way to overcome this difference. To go from 2) to back to 1), you just have to substract 9. So you can write on the left-hand side of your equation, without changing the equation,

(b-2)^2 - 9 = 16

(b-2)^2 = 25

(b1-2) = 5

(b2-2) = -5


Now put (b-2)^2=16, you get


Therefore, b=6,-2

  • $\begingroup$ Please use LaTeX instead of this programmers code... it's so ugly. $\endgroup$ – Myself Apr 1 '11 at 0:14
  • $\begingroup$ @Myself: hehe...I haven't learned latex!! $\endgroup$ – user754 Apr 1 '11 at 0:17
  • $\begingroup$ Please learn some basic LaTeX, then. $\endgroup$ – Arturo Magidin Apr 1 '11 at 3:36

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