Question

if x+ y=5 and xy = 4 find x-y using identities ​

Answer 1

Answer:

Given,

$\bf \: x + y = 5 \: \: \: and \: \: \: xy = 4 \\ \sf \: x - y = \: to \: find \: \\ \bf we \: know \: that \\ \\ \: \: \: \: \: \: \: \: \: \bf \: {(x - y)}^{2} = {(x + y)}^{2} - 4xy \\ \implies \bf \: {(x - y)}^{2} = {5}^{2} - 4 \times 4 \\ \implies \bf \: {(x - y)}^{2} = 25 - 16 = 9 \\ \implies \bf \: x - y = \sqrt{9} \\ \implies \bf \: x - y = 3$

Additional Information :-

Important identities :-

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \bf{ \: {x}^{2} - {y}^{2} = (x - y)(x + y)}}$

$\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}}$

$\boxed{ \bf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )}}$

$\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )}}$

Answer 2

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\:x + y = 5}$

and

$\red{\rm :\longmapsto\:xy = 4}$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \bf{ x \: - \: y \: using \: suitable \: identity}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x + y = 5$

and

$\rm :\longmapsto\:x y = 4$

We know,

$\red{\rm :\longmapsto\: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy - - - - (1)}$

and

$\red{\rm :\longmapsto\: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy - - - - (2)}$

On Subtracting equation (2) from equation (1), we get

$\red{\rm :\longmapsto\: {(x + y)}^{2} - {(x - y)}^{2} = {x}^{2}+{y}^{2} + 2xy - {x}^{2} - {y}^{2} + 2xy}$

$\red{\rm :\longmapsto\: {(x + y)}^{2} - {(x - y)}^{2} =4xy}$

On substituting the values of x + y and xy, we get

$\red{\rm :\longmapsto\: {(5)}^{2} - {(x - y)}^{2} =4 \times 4}$

$\red{\rm :\longmapsto\: 25 - {(x - y)}^{2} =16}$

$\red{\rm :\longmapsto\: - {(x - y)}^{2} =16 - 25}$

$\red{\rm :\longmapsto\: - {(x - y)}^{2} = - 9}$

$\red{\rm :\longmapsto\: {(x - y)}^{2} = 9}$

$\red{\rm :\longmapsto\: {x - y} = \: \pm \: 3}$

Additional Information :-

More Identities to know :-

$\blue{\rm :\longmapsto\: {(a + b)}^{2} = {(a - b)}^{2} + 4ab}$

$\blue{\rm :\longmapsto\: {(a - b)}^{2} = {(a + b)}^{2} - 4ab}$

$\blue{\rm :\longmapsto\: {(a + b)}^{3} = {a}^{3} + {3a}^{2}b + {3ab}^{2} + {b}^{3} }$

$\blue{\rm :\longmapsto\: {(a - b)}^{3} = {a}^{3} - {3a}^{2}b + {3ab}^{2} - {b}^{3} }$

$\blue{\rm :\longmapsto\:(a + b)( {a}^{2} - ab + {b}^{2}) = {a}^{3} + {b}^{3}}$

$\blue{\rm :\longmapsto\:(a - b)( {a}^{2} + ab + {b}^{2}) = {a}^{3} - {b}^{3}}$

$\blue{\rm :\longmapsto\: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}$

$\blue{\rm :\longmapsto\: {a + b + c = 0} \implies \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc }$