Question

If 2x/3y=3y/4z then show that 4x square -9y square+16z square=(2x -3y+4z) (2x+3y+4z)

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ \frac{2x}{3y} = \frac{3y}{4z} }$

TO PROVE

$\sf{4 {x}^{2} - 9 {y}^{2} + 16 {z}^{2} = (2x - 3y + 4z)(2x + 3y + 4z)\: }$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

$\sf{ {a}^{2} - {b}^{2} = (a + b)(a - b) \: }$

PROOF

Here it is given that

$\displaystyle \sf{ \frac{2x}{3y} = \frac{3y}{4z} }$

On Cross Multiplication we get

$\sf{9 {y}^{2} = 8xz }$

RHS

$\sf{ = (2x - 3y + 4z)(2x + 3y + 4z)\: }$

$\sf{ = (2x + 4z- 3y )(2x+ 4z + 3y )\: }$

$\sf{ = (2x + 4z + 3y )(2x+ 4z - 3y )\: }$

$\sf{ ={ (2x + 4z)}^{2} - {(3y)}^{2} } \: (using \: above \: identity)$

$= \sf{ 4 {x}^{2} + 16xz + 16 {z}^{2} - 9 {y}^{2} \: }$

$= \sf{ 4 {x}^{2} + 16xz + 16 {z}^{2} - 8xz\: } \: ( \because \: \sf{9 {y}^{2} = 8xz })$

$= \sf{ 4 {x}^{2} + 8xz + 16 {z}^{2} \: } \:$

$= \sf{ 4 {x}^{2} + 9 {y}^{2} + 16 {z}^{2} \: } (\because \sf{9 {y}^{2} = 8xz })$

= LHS

Hence proved

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