Radicals

Algebra Level 2

$\large \sqrt{4 + \sqrt{7}} - \sqrt{4 - \sqrt{7}} = \ ?$

\sqrt{3}

\sqrt{2}

6 solutions

Tay Yong Qiang
Aug 17, 2015

$a =$ $\sqrt{4+\sqrt{7}}-$ $\sqrt{4-\sqrt{7}}$

$a^2 =$ $4+\sqrt{7}$ $-2\sqrt{16-7}$ $+4-\sqrt{7}$

$a^2 =$ $8- 2\sqrt{9} = 2$

$a = \boxed{\sqrt{2}}$

I forgot to take the square root of a (Bangs head against wall)

John Taylor - 5 years, 9 months ago

Why did you discard $-\sqrt{2}$ ?

Yasir Soltani - 5 years, 4 months ago

In order for the radical sign to be a function it must return only one value, so the radicals in the given expression $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}$ refer to the 'principal square root' or positive square root. Knowing that, you can see that the expression must have a positive value.

Luke Johnson-Davies - 5 years, 4 months ago

It is clear that $\sqrt{4+\sqrt{7}}>\sqrt{4-\sqrt{7}}$ , i can't see how you can use the basis of Principal square root to eliminate $-\sqrt{2}$

Yasir Soltani - 5 years, 4 months ago

@Yasir Soltani – But the fact that $\sqrt{4+\sqrt{7}}>\sqrt{4-\sqrt{7}}$ means that the subtraction must have a result greater than zero, so a negative result such as $-\sqrt{2}$ is impossible.

Luke Johnson-Davies - 5 years, 4 months ago

@Luke Johnson-Davies – That was a point i was trying to make, i just didn't get your argument!

Yasir Soltani - 5 years, 4 months ago

@Yasir Soltani – Oh sorry, my point was that the only way the original expression could be negative is if you were taking (some of) the radicals to be negative square roots.

Luke Johnson-Davies - 5 years, 4 months ago

true answer

Abu Afghani - 5 years, 4 months ago

Khang Nguyen Thanh
Aug 17, 2015

$\quad\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}$

$=\dfrac{1}{\sqrt2}\left(\sqrt{8+2\sqrt7}-\sqrt{8-2\sqrt7}\right)$

$=\dfrac{1}{\sqrt2}\left(\sqrt{\left(\sqrt7+1\right)^2}-\sqrt{\left(\sqrt7-1\right)^2}\right)$

$=\dfrac{1}{\sqrt2}\left(\sqrt7+1-\left(\sqrt7-1\right)\right)$

$=\dfrac{1}{\sqrt2}. 2=\boxed{\sqrt2}$ .

Did u think it yourself

Shashank Rustagi - 5 years, 2 months ago

It's not too hard to think .. one can always think to make the thing inside the square root a perfect square :)

Anurag Pandey - 4 years, 9 months ago

Ben Habeahan
Aug 18, 2015

$\sqrt{4+\sqrt{7}} - \sqrt{4-\sqrt{7}} \\ = \sqrt{ \frac{2(4+\sqrt7)}{2}} - \sqrt{ \frac{2(4-\sqrt7)}{2}} \\ = \sqrt{ \frac{8+2 \sqrt7}{2}} - \sqrt{ \frac{8-2\sqrt7}{2}} \\ = \frac{ \sqrt{ 8+2\sqrt7}}{\sqrt{2}} - \frac{ \sqrt{ 8-2\sqrt7}}{\sqrt{2}} \cdots (1) \\$
use formula $\\ \sqrt{ (a+b) \pm 2 \sqrt{ a \times b}} = \sqrt{a} \pm \sqrt{b}$ (With $a>b>0$ ) From (1), $= \frac{ \sqrt{ (7+1)+ 2\sqrt{7 \times 1}}}{\sqrt{2}} - \frac{ \sqrt{ (7+1)- 2\sqrt{7 \times 1}}}{\sqrt{2}} \\ = \frac{ \sqrt{ 7} + \sqrt{ 1}}{ \sqrt{ 2}} - \frac{ \sqrt{ 7} - \sqrt{ 1}}{ \sqrt{ 2}} \\ = \frac{ 2}{ \sqrt{ 2}} = \boxed{ \sqrt{ 2}}$

Betty BellaItalia
Apr 15, 2017

Mohamed Wafik
Aug 21, 2015

$\sqrt{4+\sqrt{7}} - \sqrt{4-\sqrt{7}} = (\sqrt{4+\sqrt{7}} - \sqrt{4-\sqrt{7}} )\times \frac{\sqrt{4+\sqrt{7}}}{\sqrt{4+\sqrt{7}}}$

$= \frac{4+\sqrt{7} - \sqrt{16-7}} { \sqrt{4+\sqrt{7}}} = \frac{4+\sqrt{7}-3}{ \sqrt{4+\sqrt{7}}}$

$= \frac {1+\sqrt{7}}{\sqrt{4+\sqrt{7}}} = \sqrt{\frac{1+2 \times \sqrt{7} + 7}{4+ \sqrt{7}}}$

$= \sqrt{\frac {2 \times (4+\sqrt{7})}{4+\sqrt{7}}} = \sqrt{2}$

William Isoroku
Aug 18, 2015

From my math, I got 4 answers: $\pm{\sqrt{2}}, \pm{\sqrt{14}}$

By definition, the square root of some real number is ever positive.

$\sqrt{4} = x$ , here $x = 2$ and just 2 because every function must return just one value.

it is not the same than $4 = x^2$ , here $x$ can be $+2$ or $-2$

because of that, $\sqrt{x^2} = |x|$ and not just $x$

Sorry for my poor English

Victor Duarte da Silva - 5 years, 9 months ago

Radicals

6 solutions

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