$\sqrt{27+5\sqrt{8}}$

Algebra Level 2

If $\sqrt{27+5\sqrt{8}}=a+\sqrt{b},$ where $a$ is an integer and $b$ is a prime number, what is $b?$

3 7 5 2

4 solutions

Damiann Mangan
Mar 12, 2014

$\sqrt{27+5 \sqrt{8}} = \sqrt{5^{2} + 2 * 5 * \sqrt{2} + \sqrt{2}^{2}} = 5 + \sqrt{2}$

which means $b = 2$ .

the answer must be +-2 because in the square root the 2 can be + or _

hassam rind - 7 years, 3 months ago

I think you are confused, he is trying to make the from $a^2 + 2ab + b^2$

Now as you are saying , $2 = (\sqrt{2})^2 ~or~ (-\sqrt{2})^2$ , it absolutely correct ,

but see what sign we are having in the form 2ab , positive?

Now suppose you say , we can make the sign of 2ab positive by take $(-5)^2 = 25$

Ok , then our final answer will be $\sqrt{( - 5 - \sqrt{2})^2} = \sqrt{(-1)^2 (5 + \sqrt{2})^2} = | -1( 5 + \sqrt{2})|$

Note - Square root of a value means its magnitude

U Z - 6 years, 5 months ago

Toby M
Jun 21, 2020

Squaring both sides, we get $27 + 5 \sqrt8 = a^2 + 2a \sqrt{b} + b$ . Equating the surd (radical) parts, we get that $5 \sqrt8 = 2a \sqrt b$ . However, $\sqrt8 = \sqrt{2^2 \times 2} = 2 \sqrt{2}$ , so $5(2\sqrt2)= 2a \sqrt{b} \Rightarrow 5 \sqrt2 = a \sqrt{b}$ . Therefore $b = \boxed{2}$ .

Maths Perera
Apr 8, 2014

U don't need to work out the whole equation. The first step, and only step, would be to simplify 5 root 8. That becomes 10 root 2. 2 is your answer (b)

Sarthak Rapartiwar
Mar 12, 2014

sqrt(27+5sqrt(8))=a+sqrt(b).......... squaring both side................ 27+5sqrt(8)=a2+b+2asqrt(b).......... ==>>it gives ............ 27+10sqrt(2)=a2 +b+2zsqrt(b).............. comparing both side................ a2+b=27 and 2asqrt(b)=10sqrt(2)................ which gives .... a=25 and b=2 Ans.

good idea, but a little hard to read... by the way, a=5, not 25

Harshal Sheth - 7 years, 2 months ago

27 + 5 8 \sqrt{27+5\sqrt{8}} 2 7 + 5 8 ​ ​

4 solutions

0 pending reports

$\sqrt{27+5\sqrt{8}}$