Compound Surds

Algebra Level 1

Using the fact that ( x + y ) 2 = x + y + 2 x y (\sqrt{x}+\sqrt{y})^2=x+y+2\sqrt{xy} , find the square root of 5 + 24 5+\sqrt{24} .
This number can be expressed in the form a + b \sqrt{a}+\sqrt{b} , where a ≤ b a\leq{b} .

Find the value of b − a b-a .

3 0 1 2

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

We may simplify the expression 5 + 24 5+\sqrt{24} into 5 + 2 6 5+2\sqrt6 . Using the fact given, ( a + b ) 2 = a + b + 2 a b ⟹ a + b = 5 , a b = 6 (\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}\implies{a+b=5, ab=6} . Noting that a ≤ b a\leq{b} , solving this system yields a = 2 , b = 3 a=2, b=3 . ∴ b − a = 1 \therefore{b-a=1}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Why not convert 5 5 to 25 \sqrt{25} Don't need to mess up with formulas.

Akshay Krishna - 2 years, 5 months ago

noooooooooooooooooooo k

Math Master - 10 months, 1 week ago
Ashish Menon
Apr 13, 2016

5 + 24 = 2 + 3 + 4 × 6 = ( 2 ) 2 + ( 3 ) 2 + 2 6 = ( 2 ) 2 + ( 3 ) 2 + 2 3 × 2 = ( 2 + 3 ) 2 = 2 + 3 \sqrt{5 + \sqrt{24}}\\ = \sqrt{2 + 3 + \sqrt{4 × 6}}\\ = \sqrt {{(\sqrt{2})}^2 + {(\sqrt{3})}^2 + 2\sqrt{6}}\\ = \sqrt {{(\sqrt{2})}^2 + {(\sqrt{3})}^2 + 2\sqrt{3}×\sqrt{2}}\\ = \sqrt {{(\sqrt{2} + \sqrt{3})}^2}\\ = \sqrt{2} + \sqrt{3}

∴ \therefore The answer is ( 3 − 2 ) = 1 (3-2) = \boxed{1}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

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