Roots

Algebra Level 3

$\large \sqrt{45 - \sqrt{2000}}+\sqrt{45 + \sqrt{2000}} =\, ?$

5

10

2\sqrt{20}

20

1 solution

Naren Bhandari
Feb 9, 2017

$\sqrt{45 - \sqrt{100.20}}$ $+$ $\sqrt{45 + \sqrt{100.20}}$

$=\sqrt{5^2 + 20 - 10\sqrt{20}}$ $+$ $\sqrt{5^2 + 20 + 10\sqrt{20}}$

$= \sqrt{(5 -\sqrt{20})^2}$ $+$ $\sqrt{(5 + \sqrt{20})^2 }$

$= 5 -\sqrt{20} + 5 + \sqrt{20}$ $=\boxed{10}$

$OR$

Let, $A = \sqrt{45 - \sqrt{2000}} + \sqrt{45 +\sqrt{2000}}$

squaring on both sides we get

$A^2 = 90 + 2\sqrt{45^2 - 2000}$

$A^2 = 90 + 10$

$A^2 = 100$

$A = \boxed{10}$

I have problems with this solution: first of all why does $\sqrt(45)$ equal $5^2+\sqrt(20)$ ? Next to that: the left root is negative so should imply an imaginary result.

Peter van der Linden - 4 years, 4 months ago

Won't let me edit: I ment the left root in the first step made by you.

Peter van der Linden - 4 years, 4 months ago

$|\sqrt{20}-\sqrt{25}|$ should give a positive value

Provided √(20 - 10√20 + 25) + √(20 +10√20 +25) =√(√20)² - 10.√20 + 5²) + √(√20 +5)² = (√20 -5) + (√20 +5)

since √20 -5 = √(−0.527864) => (imaginary) (which isnot appropiate)

Naren Bhandari - 4 years, 4 months ago

That's what I tried to say :)

Peter van der Linden - 4 years, 4 months ago

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