[SOLVED] Expressing in the form?

**mark1950** · June 12th 2009, 06:53 PM

Express $\sqrt{44-24\sqrt{2}}$ in the form $a + b\sqrt{2}$

**Soroban** · June 12th 2009, 07:40 PM

Hello, Mark!

Express $\sqrt{44-24\sqrt{2}}$ in the form $a + b\sqrt{2}$

$\text{We have: }\;\sqrt{44-24\sqrt{2}} \;=\;a + b\sqrt{2}\;\;\text{ where }a\text{ and }b\text{ are rational numbers.}$

Square both sides: . $44 - 24\sqrt{2} \:=\:(a+b\sqrt{2})^2 \:=\:a^2 + 2ab\sqrt{2} + 2b^2$

And we have: . $(a^2 + 2b^2) + (2ab)\sqrt{2} \;=\;44-24\sqrt{2}$

Equate coefficients: . $\begin{array}{ccc}a^2+b^2 \:=\:44 & {\color{blue}[1]} \\ 2ab \:=\:-24 & {\color{blue}[2]} \end{array}$

From ${\color{blue}[2]}\!:\;\;b \:=\:-\frac{12}{a}\;\;{\color{blue}[3]}$

$\text{Substitute into }{\color{blue}[1]}\!:\;\;a^2 + 2\left(-\frac{12}{a}\right)^2 \:=\:44 \quad\Rightarrow\quad a^2 + \frac{288}{a^2} \:=\:44<br />$

Multiply by $a^2\!:\;\;a^4 + 288 \:=\:44a^2 \quad\Rightarrow\quad a^4 - 44a^2 + 288 \:-\:0$

Factor: . $(a^2 - 36)(a^2-8) \:=\:0 \quad\Rightarrow\quad (a-6)(a+6)(a^2-8) \:=\:0$

. . And the rational roots are: . $a \:=\:\pm 6$

Substitute into ${\color{blue}[3]}\!:\;\;b \:=\:-\frac{12}{\pm6} \:=\:\mp 2$

There are two solutions: . $a + b\sqrt{2}\;=\;\begin{Bmatrix}6 - 2\sqrt{2} \\ \text{-}6 + 2\sqrt{2} \end{Bmatrix}$

**mark1950** · June 13th 2009, 02:08 AM

What's the difference between $\pm$ and $\mp$ ? Why do you have to inverse them to get $\mp2$ ? Isn't it okay to just put $\pm2$ ?

**yeongil** · June 13th 2009, 02:13 AM

Originally Posted by mark1950

What's the difference between $\pm$ and $\mp$ ? Why do you have to inverse them to get $\mp2$ ? Isn't it okay to just put $\pm2$ ?

No. Looking at this from Soroban's post:
$b \:=\:-\frac{12}{\pm6} \:=\:\mp 2$

See the $\pm6$ in the denominator? The $\mp$ sign indicates that the number that follows has to be the opposite of whatever sign the number after the $\pm$ is. If the denominator is +6, then it simplifies to -2, and if the denominator is -6, it simplifies to +2.

The $\mp$ isn't used as often as $\pm$ . If you know trig, then you may have seen the sum/difference identity for cosine combined into one identity like this:
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ .

Of course, the two identities are these:
$\cos(A + B) = \cos A \cos B - \sin A \sin B$ .
$\cos(A - B) = \cos A \cos B + \sin A \sin B$ .

Note how the sign changes on the right side. That's the reason we use the $\mp$ sign.

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