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"Write the following expressions in the form $X^2 − A^2$ using the method of completing the square."

"1) $x^2 + 8x + 9$"

I don't understand how that can be, but I did try to do it, with no obvious next steps.

My solution:-

$x^2 + 8x + 16 -16 + 9$

$(x + 4)^2 - 7$

I completed the square, but how can I put that in the form $X^2 - A^2$?

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    $\begingroup$ Can you think of any numbers $A$ such that $A^2=7$? $\endgroup$ – Zev Chonoles Mar 4 '12 at 23:08
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    $\begingroup$ Note that $X$ is not the same thing as $x$. (Yes, case matters!). So, we can let $X$ be $x+4$; that takes care of the "$X^2$" part. For the $A^2$ part, use Zev's hint. $\endgroup$ – Arturo Magidin Mar 4 '12 at 23:10
  • $\begingroup$ I took the liberty of changing the $4x$ to $8x$ as the rest of the work seems to suggest you did, in fact, mean $8x$ $\endgroup$ – Juan S Mar 4 '12 at 23:12
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    $\begingroup$ If you understand, write out your understanding as an answer. If after a few days no one has pointed out any mistakes, you can accept your answer. $\endgroup$ – Gerry Myerson Mar 4 '12 at 23:54
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    $\begingroup$ See <a href="youtube.com/watch?v=_GBtlR4m67g&feature=relmfu"> this video </a> for example for completing the square. $\endgroup$ – Scott Carter Mar 5 '12 at 0:15
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'$x$' isn't the same as '$X$', so $(x+4)^2 − 7$ can be written in the form $X^2 - A^2$. Particularly, $(x+4)$ can be considered as $X$, and $\sqrt{7}$ can be considered as $A$. So, the expression in the designated form would be $(x+4)^2 - (\sqrt{7})^2$.

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In general, given $ax^2 +bx + c\ \forall x \in \mathbb R$ where $a \ne 0$.$$ax^2+bx+c= X^2- A^2$$ where $$X= x+\frac{b}{2a} $$ and $$A= \frac {\sqrt{b^2-4ac}}{2a}$$. Using the quadratic formula and some algebraic mapulation we have the above result

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