0
$\begingroup$

I am trying to put the following expression $x^2 + 6x + 7$ in the form $(x + r)^2 + s$ so I can complete the square.

I understand the first couple of steps,

$(x^2 + 6x + 7)$

$b =\left(\frac62\right)^2 = 9$

$x^2 + 6x + 9 - 9 + 7$

$(x^2 + 6x + 9) - 9 + 7$

$(x^2 + 6x + 9) - 2$

From this point on I'm not sure what to do to put it in $(x + r)^2 $. What is $r?$ Any help is much appreciated.

Also, I have one question last important question:

1) Will this method work for any expression / equation in solving the square?

$\endgroup$
  • $\begingroup$ Write $x^2+6x+9$ as the square of a linear term. $\endgroup$ – David Mitra Jan 18 '14 at 11:37
0
$\begingroup$

Note that $$x^2+2ax+a^2=(x+a)^2.$$

So, in your case, $$x^2+6x+9=x^2+2\cdot 3x+3^2=(x+3)^2.$$

In general, $$\begin{align}ax^2+bx+c&=a\left(x^2+\frac{b}{a}x\right)+c\\&=a\left\{x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right\}+c\\&=a\left\{\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right\}+c\\&=a\left(x+\frac{b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c.\end{align}$$

$\endgroup$
  • $\begingroup$ do you mean $r=-3$? $\endgroup$ – nadia-liza Jan 18 '14 at 11:39
  • $\begingroup$ Sorry but how did you get r? $\endgroup$ – Sophia Jan 18 '14 at 11:39
  • $\begingroup$ @Sophia: $r=6/2$. Note that $x^2+2ax+a^2=(x+a)^2$. In other word, $r$ is the half of the coefficient of $x$. $\endgroup$ – mathlove Jan 18 '14 at 11:46
  • $\begingroup$ @nadia-liza: No. I think $r=3$. $\endgroup$ – mathlove Jan 18 '14 at 11:55
1
$\begingroup$

If you don't understand / remember the algorithm, then ignore the algorithm and simply solve for things.

You want

$$ x^2 + 6x + 7 = (x+r)^2 + s$$

so you expand the right hand side and solve for values of $r$ and $s$ that make the two sides equal.

$\endgroup$
0
$\begingroup$

I was taught you half the x coefficient and subtract its square. Eg:

$$ x^2+6x+7 = (x+3)^2 - (3)^2+7 = (x+3)^2-2 $$

Or in general, $$x^2+bx+c=(ax+\frac{b}{2})^2-(\frac{b}{2})^2+c$$

This works in all cases. If b is negative, then you would get: $$x^2-bx+c=(x-\frac{b}{2})^2-(\frac{b}{2})^2+c$$

Or if the term in front of $x^2$ was a, take a out as a factor and continue from there: $ax^2+bx+c = a(x^2+\frac{bx}{a}+\frac{c}{a})$

$\endgroup$
0
$\begingroup$

You surely know that $(a+b)^2=a^2+2ab+b^2$. Now your problem is to transform $(x^2 + 6x + 7)$ in the form $(x+r)^2+s$ right? We can notice that $(x^2+6x)$ is similar to $a^2+2ab$ in the expansion of $(a+b)^2$ but to complete the expansion we need a value that multiplied by $(2\cdot x)$ give us $6x$ and that value is 3. To complete the expansion we need to square the value (that gives us 9). So we have$$x^2 + 6x + 7=x^2 + 6x + 9 - 9 + 7 = (x^2+6x+9)-2= (x+3)^2-2$$ Your equation written in the form $(x+r)^2+s$

Hope this will help

peterix

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.