How do you simplify $(4 + \sqrt{2}) \div (8 - \sqrt{2})$ $(4+sqrt2 )div (8 - sqrt2)$ ?

Question

How do you simplify $(4 + \sqrt{2}) \div (8 - \sqrt{2})$ $(4+sqrt2 )div (8 - sqrt2)$ ?

1 Answer

Impact of this question

2066 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-20T19:40:42

$\frac{17 + 6 \sqrt{2}}{31}$ $(17+6sqrt2)/31$

Explanation:

Let's rewrite this in the fractional form:
$\frac{4 + \sqrt{2}}{8 - \sqrt{2}}$ $(4+sqrt2)/(8-sqrt2)$

Then, I would multiply, top and bottom, by $(8 + \sqrt{2})$ $(8+sqrt2)$ so that:
$\frac{4 + \sqrt{2}}{8 - \sqrt{2}} \cdot \frac{8 + \sqrt{2}}{8 + \sqrt{2}}$ $(4+sqrt2)/(8-sqrt2) * (8+sqrt2)/(8+sqrt2)$

(Note that $\frac{8 + \sqrt{2}}{8 + \sqrt{2}} = 1$ $(8+sqrt2)/(8+sqrt2)=1$ , so multiplying by this factor does not change the results at all).

I chose $(8 + \sqrt{2})$ $(8+sqrt2)$ because
$(a - b) \cdot (a + b) = a^{2} - b^{2}$ $(a-b)*(a+b)=a^2-b^2$
and I am hoping it will make the solution easier.

Then the fraction becomes:
$\frac{(4 + \sqrt{2}) (8 + \sqrt{2})}{8^{2} - {(\sqrt{2})}^{2}} = \frac{32 + 12 \sqrt{2} + 2}{64 - 2}$ $((4+sqrt2)(8+sqrt2))/(8^2 - (sqrt2)^2) = (32+12sqrt2+2)/(64-2)$
which is
$= \frac{34 + 12 \sqrt{2}}{62}$ $=(34+12sqrt2)/(62)$
We can factor out 2 from the top and bottom:
$= \frac{2 \cdot (17 + 6 \sqrt{2})}{2 \cdot 31}$ $=(2*(17+6sqrt2))/(2*31)$
$= \frac{17 + 6 \sqrt{2}}{31}$ $=(17+6sqrt2)/31$

Science

Math

Humanities

... and beyond

How do you simplify $(4 + \sqrt{2}) \div (8 - \sqrt{2})$ $(4+sqrt2 )div (8 - sqrt2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you simplify (4+√2)÷(8−√2)(4+sqrt2 )div (8 - sqrt2)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you simplify $(4 + \sqrt{2}) \div (8 - \sqrt{2})$ $(4+sqrt2 )div (8 - sqrt2)$ ?