Algebraic trouble - No.1

Algebra Level 1

If a + b = 5 a + b = 5 and a b = 3 a - b = 3 , then find the value of the following expression: 16 a b ( a 2 + b 2 ) 16ab(a^2 + b^2)


The answer is 1088.

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2 solutions

Note: This is a very Basic approach which is expected to be useful for beginners.

We can simply separate this expression into 3 sections:

S1: Finding a 2 + b 2 a^2 + b^2

S2: Finding 4 a b 4ab

S3: It is just about solving the expression

So lets begin with: Section 1

To find the values of a 2 + b 2 a^2 + b^2 we use the formula:

( a + b ) 2 + ( a b ) 2 = 2 ( a 2 + b 2 ) (a + b)^2 + (a - b)^2 = 2(a^2 + b^2)

Putting the values of a + b a + b and a b a - b , we get:

2 ( a 2 + b 2 ) = ( 5 ) 2 + ( 3 ) 2 = 25 + 9 = 34 2(a^2 + b^2) = (5)^2 + (3)^2= 25 + 9 = 34 or

( a 2 + b 2 ) (a^2 + b^2) = 34 2 ) \frac{34}{2)} = 17

Section 2

To find the value of 4 a b 4ab we use the formula:

( a + b ) 2 ( a b ) 2 = 4 a b (a + b)^2 - (a - b)^2 = 4ab

** I'm lazy to type. just do the calculations since you know the formula

You are sure to get:

4 a b = 16 4ab = 16

Section 3

To find the value of 16 a b ( a 2 + b 2 ) 16ab(a^2 + b^2) , we have

16 a b ( a 2 + b 2 ) = 4 ( 4 a b ) ( a 2 + b 2 ) 16ab(a^2 + b^2) = 4(4ab)(a^2 + b^2)

Using the result in S1 and S2

= 4 (16) (17)

= 1088

Kenny O.
Sep 1, 2017

(a+b)-(a-b)=5-3 = > => 2b=2 = > => b=1
a+1=5 = > => a=4
16ab( a 2 + b 2 a^2+b^2 )=16(4)(1)( 4 2 + 1 2 4^2+1^2 )=1088

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