Question

if x+y =4 and xy=2 find x²+y²​

Answer 1

Answer

• x² + y² = 12

Given

• x + y = 4 and xy = 2

To Find

• The value of x² + y²

Step By Step Explanation

It is given that => x + y = 4 and xy = 2

We need to find the value of x² + y²

So let's do it !!

Using Identity ⤵

$\maltese\boxed{ \tt{ \red{{(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

By substituting the values

$\longmapsto \sf{(4)}^{2} = {x}^{2} + {y}^{2} + 2(2) \\ \\ \longmapsto \sf 16 = {x}^{2} + {y}^{2} + 4 \\ \\ \longmapsto \sf16 - 4 = {x}^{2} + {y}^{2} \\ \\ \longmapsto {\bf{ \green{12 = {x}^{2} + {y}^{2} }}}$

Therefore, x² + y² = 12

♠ Extra Identities

• x² - y² = ( x - y ) ( x + y )

• ( x + y )² = x² + y² + 2xy

• ( x - y )² = x² + y² - 2xy

• ( x + a ) ( x + b ) = x² + x( a + b ) + ab

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Answer 2

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$\sf \small \red{given : }$

$\: \: \: \: \: \: \sf\cdot \: x+y =4 \: and \: xy=2$

$\sf\small\red{Find:-}$

$\: \: \: \: \: \: \: \ \sf \cdot \: Find \: {x}^{2} + {y}^{2}$

$\sf\small\red{Identify:-}$

$\: \: \: \sf \: \: \: \cdot {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

$\sf\small\red{Solution:-}$

$\sf\leadsto {(4)}^{2} = {x}^{2} + {y}^{2} + 2(2)$

$\sf\leadsto 16 = {x}^{2} + {y}^{2} + 2(2)$

$\sf\leadsto {x}^{2} + {y}^{2} = 16 - 4$

$\dashrightarrow \sf\ {x}^{2} + {y}^{2} = 12 \dashleftarrow$

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