Question

if 3x + 4y = 16 and xy=4 ; find the value of 9x^2+16y^2

Answer 1

[0010001111]... Hello User... [1001010101]
Here's your answer...

3x+4y = 16
xy = 4
The identity we will use here is...
(a+b)² = a² + b² + 2ab
(3x+4y)² = (3x)² + (4y)² + 2×3x×4y
(3x+4y)² = 9x² + 16y² + 24xy
16² = 9x² + 16y² +24×4
256 = 9x² + 16y² + 96
9x² + 16y² = 256 - 96 = 160

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Answer 2

Hey Sweta !

Here is your solution :

Given,

3x + 4y = 16 , xy = 4.

Now ,

( A + B )^2 = A^2 + B^2 + 2 AB

( 3x +4y )^2=(3x)^2 + ( 4y )^2 + 2 × 3x × 4y

( 3x + 4y)^2 = 9x^2 + 16y^2 + 24 xy

By putting the value of ( 3x + 4y ) and ( xy ).


( 16 )^2 = 9x^2 + 16y^2 + 24 × 4

256 = 9x^2 + 16y^2 + 96

256 - 96 = 9x^2 + 16y^2

160 = 9x^2 + 16y^2.

The required answer is 160.


Hope it helps !!