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Algebra Level 2

If a 2 + b 2 = 4 a b a^2+b^2=4ab what is ( a + b a − b ) 2 \left(\frac{a+b}{a-b}\right)^2 ?


The answer is 3.

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Daniel Lombardo by Daniel Lombardo , 20, Mexico

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8 solutions

Victor Loh
Jul 8, 2014

Note that

( a + b a − b ) 2 = a 2 + b 2 + 2 a b a 2 + b 2 − 2 a b \left(\frac{a+b}{a-b}\right)^2=\frac{a^2+b^2+2ab}{a^2+b^2-2ab}

Since a 2 + b 2 = 4 a b a^2+b^2=4ab , we have

a 2 + b 2 + 2 a b a 2 + b 2 − 2 a b = 4 a b + 2 a b 4 a b − 2 a b \frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{4ab+2ab}{4ab-2ab}

= 6 a b 2 a b = 3 . =\frac{6ab}{2ab}=\boxed{3}.

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Note that this is only true if both a and b are not 0

Marco Ceze - 4 years, 5 months ago
Finn Hulse
Apr 6, 2014

First expand what we want to find: a 2 + 2 a b + b 2 a 2 − 2 a b + b 2 . \frac{a^{2}+2ab+b^{2}}{a^{2}-2ab+b^{2}}. Let's reexamine the first equation that we're given. To complete the square, let's add 2 a b 2ab to both sides. The equation can now be factored as ( a + b ) 2 = 6 a b . (a+b)^{2}=6ab. Thus, plugging this back into the expansion of the second equation, we produce 6 a b ( a − b ) 2 . \frac{6ab}{(a-b)^{2}}. Now to find a clever way to solve for the denominator. Subtract 2 a b 2ab from the first equation. Factor as ( a − b ) 2 = 2 a b . (a-b)^{2}=2ab. By substitution, the second equation is now 6 a b 2 a b = 3 . \frac{6ab}{2ab} =\boxed{3}.

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Moderator note:

Great! What would the value of a 3 − 3 a 2 b + 3 a b 2 − b 3 a 3 + 3 a 2 b + 3 a b 2 + b 3 \frac {a^3-3a^2b + 3ab^2 - b^3}{a^3+3a^2 b + 3ab^2+b^3} be?

Your question would turn into (a-b)^2 (a-b)/(a+b)^2 (a+b) =2ab sqrt(2ab) / 6ab sqrt(6ab) =1/3*sqrt3 where sqrt is square root of the figure. Sorry for the lag I just joined 'Brilliant'.

Swapnil Joshi - 6 years, 1 month ago

Challenge student's note: The given expression can be simplified as follow:

a 3 − 3 a 2 b + 3 b 2 a − b 3 a 3 + 3 a 2 b + 3 b 2 a + b 3 = ( a − b ) 3 ( a + b ) 3 \frac { { a }^{ 3 }-3{ a }^{ 2 }b+3{ b }^{ 2 }a-{ b }^{ 3 } }{ { a }^{ 3 }+3{ a }^{ 2 }b+3{ b }^{ 2 }a+{ b }^{ 3 } } =\frac { { (a-b) }^{ 3 } }{ { (a+b) }^{ 3 } }

Since ( a + b ) 2 = 6 a b { (a+b) }^{ 2 }=6ab ,

a+b=\sqrt { 6ab } \\ { (a+b) }^{ 3 }={ (\sqrt { 6ab } ) }^{ 3 }\\ { (a+b) }^{ 3 }=6ab{ \sqrt { 6ab }  }

Again since ( a − b ) 2 = 2 a b { (a-b) }^{ 2 }=2ab ,

a − b = 2 a b ( a − b ) 3 = 2 a b 2 a b { a-b }=\sqrt { 2ab } \\ { (a-b) }^{ 3 }=2ab\sqrt { 2ab }

On substituting these values in the equation, we get

2 a b 2 a b 6 a b 6 a b = 1 3 1 3 = 1 3 3 \frac { 2ab\sqrt { 2ab } }{ 6ab\sqrt { 6ab } } =\frac { 1 }{ 3 } \sqrt { \frac { 1 }{ 3 } } =\frac { 1 }{ 3\sqrt { 3 } }

Arulx Z - 6 years ago

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A more definite answer would be sqrt(3)/9, since it is hard to divide by irrationals. 2ab sqrt(2ab)/6ab sqrt(6ab) can be simplified into: sqrt(2)/3 sqrt(6) =sqrt(12)/18 =2 sqrt(3)/18 =sqrt(3)/9

Ryan Neal - 4 years, 7 months ago
Daniel Lombardo
Apr 6, 2014

(a+b/a-b)^2 = (a+b)^2/(a-b)^2= a^2+2ab+b^2/a^2-2ab+b^2= 4ab+2ab/4ab-2ab=6ab/2ab=3

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Arulx Z
May 26, 2015

a 2 + b 2 = 4 a b a 2 + 2 a b + b 2 = 6 a b ( a + b ) 2 = 6 a b a 2 − 2 a b + b 2 = 2 a b ( a − b ) 2 = 2 a b ( a + b ) 2 ( a − b ) 2 = 6 a b 2 a b = 3 { a }^{ 2 }+{ b }^{ 2 }=4ab\\ \\ { a }^{ 2 }+2ab+{ b }^{ 2 }=6ab\\ { (a+b) }^{ 2 }=6ab\\ \\ { a }^{ 2 }-2ab+{ b }^{ 2 }=2ab\\ { (a-b) }^{ 2 }=2ab\\ \\ \frac { { (a+b) }^{ 2 } }{ { (a-b) }^{ 2 } } =\frac { 6ab }{ 2ab } =3

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Best response right here. Though longer, notating the very simple operations of +2ab and -2ab to the equations would make how simple this solution is much clearer.

Blood Blood - 4 years, 6 months ago
Isaac Jiménez
May 10, 2014

First, we know a 2 + b 2 = 4 a b ⇒ a 2 − 2 a b + b 2 = 2 a b ⇒ ( a − b ) 2 = 2 a b { a }^{ 2 }+{ b }^{ 2 }=4ab\Rightarrow { a }^{ 2 }-2ab+{ b }^{ 2 }=2ab\Rightarrow { (a-b) }^{ 2 }=2ab . Next, ( a + b ) ( a − b ) 2 = ( ( a + b ) 2 ( a − b ) 2 ) = ( ( a + b ) 2 2 a b ) = a 2 + 2 a b + b 2 2 a b = a 2 2 a b + 2 a b 2 a b + b 2 2 a b = a 2 + b 2 2 a b + 1 = 4 a b 2 a b + 1 = 2 + 1 = 3 \frac { (a+b) }{ (a-b) } ^{ 2 }=\left( \frac { { (a+b) }^{ 2 } }{ { (a-b) }^{ 2 } } \right) =\left( \frac { (a+b)^{ 2 } }{ 2ab } \right) =\frac { { a }^{ 2 }{ +2ab }{ +b }^{ 2 } }{ 2ab } =\frac { { a }^{ 2 } }{ 2ab } +\frac { 2ab }{ 2ab } +\frac { { b }^{ 2 } }{ 2ab } =\frac { { a }^{ 2 }{ +b }^{ 2 } }{ 2ab } +1=\frac { 4ab }{ 2ab } +1=2+1=3 . Remember a 2 + b 2 = 4 a b { a }^{ 2 }+{ b }^{ 2 }=4ab , so the answer is 3 \boxed { 3 }

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Joshua Nesseth
Oct 23, 2016

I don't know if it's the "best" way, but this was sure a lot simpler (for me, at least) than the routes I see here:

If we expand the numerator and denominator of the rational formula, we can easily see that the first equation can easily be put into terms of either; the denominator, however, has the added benefit of reducing the value of the coefficient.

Thus, a 2 − 2 a b + b 2 = 2 a b a^2 - 2ab + b^2 = 2ab can be substituted for the denominator; when rearranged and "simplified", this gives us a 2 + b 2 2 a b + 2 a b 2 a b \frac {a^2 + b^2}{2ab} + \frac {2ab}{2ab} .

If we allow this to equal a dummy variable (say, "y"), we can see y = a 2 + b 2 2 a b + 1 y = \frac {a^2 + b^2}{2ab} + 1 ; OR ( y − 1 ) ( 2 a b ) = a 2 + b 2 (y - 1)(2ab) = {a^2 + b^2} . We can now substitute the original equation in again (though this time unmodified), giving us ( y − 1 ) ( 2 a b ) = 4 a b (y - 1)(2ab) = 4ab , which further simplifies into ( y − 1 ) = 2 [ a b a b ] (y-1) = 2[\frac {ab}{ab}] .

Solving for y, we have y = 2 ( 1 ) + 1 y = 2(1) + 1 .

Again, perhaps not as "clean" as the other answers, but (again, for me at least) it was far less daunting than the explicit calculations--then again, I'm pretty lazy.

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Jolene :)
Apr 29, 2015

Since \­( a^2 + b^2 = 4ab ), we can manipulate the equation to get \­( a^{2} = 4ab - b^{2} )

Expand the fraction \­( ( \frac{a+b}{a-b})^{2} to get \frac{a^{2} +2ab + b^{2}}{a^{2} - 2ab + b^{2}} )

Substitute \­( a^{2)} with \­(4ab - b^{2 }) in the equation and simplify - then you'd get the answer of 3! :)

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Jrenz Salvador
May 11, 2014

through expansion i get 3 as an answer.

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