a and b

Algebra Level 2

If a a and b b are real numbers such that a b = 8 ab = 8 and a + b = 10 a + b = 10 , what is the value of a 2 + b 2 a^ 2 + b^ 2 ?


The answer is 84.

Sara Sharma by sara sharma , 18, India

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8 solutions

Harsh Shrivastava
Aug 17, 2014

( a + b ) 2 = a 2 + b 2 + 2 a b (a + b)^{2} = a^{2}+b^{2}+2ab

Plugging in the given values, we get the answer as 84 \boxed{84} .

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But what is the value of a and b??

Anuj Shikarkhane - 6 years, 9 months ago

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I know you want to find the value of a a and b b to complete this problem, although we don't need to find it. Here is the solution:

First, according to the problem, we got:

{ a ≠ 0 b ≠ 0 \begin{cases} a\neq 0 \\ b\neq 0 \end{cases}

We have:

a b = 8 ⇒ a = 8 b ab=8\Rightarrow a=\frac { 8 }{ b }

So:

a + b = 10 ⇔ a + 8 a = 10 a+b=10\Leftrightarrow a+\frac { 8 }{ a } =10

Multiply all the two sides of this equation with a a , we have:

a 2 + 8 = 10 a ⇔ a 2 − 10 a + 8 = 0 { a }^{ 2 }+8=10a\Leftrightarrow { a }^{ 2 }-10a+8=0

Solving this equation, we receive 2 2 values of a a :

a 1 = 5 + 17 { a }_{ 1 }=5+\sqrt { 17 }

a 2 = 5 − 17 { a }_{ 2 }=5-\sqrt { 17 }

Using the values of a to the equation, we receive 2 2 values of b b

b 1 = 5 − 17 { b }_{ 1 }=5-\sqrt { 17 }

b 2 = 5 + 17 { b }_{ 2 }=5+\sqrt { 17 }

So there are 2 2 pairs of ( a , b ) (a,b) :

( a , b ) = ( 5 + 17 , 5 − 17 ) (a,b)=(5+\sqrt { 17 } ,5-\sqrt { 17 })

( a , b ) = ( 5 − 17 , 5 + 17 ) (a,b)=(5-\sqrt { 17 } ,5+\sqrt { 17 })

Dang Anh Tu - 6 years, 9 months ago

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Okay..... Thanks

Anuj Shikarkhane - 6 years, 9 months ago

But what is the need of finding value of a a & b b @Anuj Shikarkhane ?

Harsh Shrivastava - 6 years, 9 months ago

a = 5 + 17 a = 5 + \sqrt{17} and b = 5 − 17 b = 5 - \sqrt{17}

Rhoy Omega - 6 years, 9 months ago

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But how did you get that??

Anuj Shikarkhane - 6 years, 9 months ago

( a + b ) 2 = 1 0 2 → a 2 + b 2 = 100 − 2 a b = 100 − 2 ( 8 ) = 100 − 16 = 84 (a+b)^2=10^2\rightarrow a^2+b^2=100-2ab=100-2(8)=100-16=\boxed{84}

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Sagar Dev
Aug 27, 2014

a 2 + b 2 = ( a + b ) 2 − 2 a b { a }^{ 2 }+{ b }^{ 2 }={ (a+b) }^{ 2 }-2ab Now Substitute the given values and you will get 84 \boxed{84}

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Ashish Menon
May 28, 2016

( a + b ) 2 = a 2 + b 2 + 2 a b ( 10 ) 2 = a 2 + b 2 + 2 × 8 a 2 + b 2 = 100 − 16 = 84 \begin{aligned} {(a + b)}^2 & = a^2 + b^2 + 2ab\\ {(10)}^2 & = a^2 + b^2 + 2×8\\ a^2 + b^2 & = 100 - 16\\ & = \color{#69047E}{\boxed{84}} \end{aligned}

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Dhruv Tyagi
May 29, 2015

Given=>a+b =10 (Equation 1) and, ab = 8 (Equation 2) What we have to find is a 2 + b 2 a^{2}+b^{2}

Using the identity ( a + b ) 2 (a+b)^{2} We get a 2 + b 2 + 2 a b = ( a + b ) 2 a^{2}+b^{2}+2ab = (a+b)^{2} then, ( a + b ) 2 − 2 a b = a 2 + b 2 (a+b)^{2}-2ab = a^{2}+b^{2} [subtracting 2ab from the simplification of the identity] ( a + b ) 2 − 2 a b = a 2 + b 2 (a+b)^{2}-2ab = a^{2}+b^{2} [we get the required solution on solving this] ( 10 ) 2 − 2 ( 8 ) = a 2 + b 2 (10)^{2} - 2(8) = a^{2}+b^{2} [using the equations (1) and (2) in the substitution method] 100-16 = a 2 + b 2 a^{2}+b^{2}

84 = a 2 + b 2 a^{2}+b^{2}

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Vishal S
Dec 16, 2014

We can write a^2+b^2 as (a+b)^2-2ab

and given ab=8 and a+b=10

By substituting the given values in a^2+b^2=(a+b)-2ab

we get a^2+b^2=10^2-2 X 8

=>100-16=84

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Ritam Baidya
Dec 3, 2014

very easy just whole sq. a+b=10 to get a 2+b 2= 100-16 = 84

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Lewis Tough
Aug 28, 2014

Squaring (a+b) gives a 2 a^{2} + b 2 b^{2} +2ab. We know ab=8, so 2ab=16. Because a+b=10 and we have squared it, we find that a 2 a^{2} + b 2 b^{2} +16 is 100. 100-16=84

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