Answer

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Hint:In this question we simply solve the expression $x + y = 36$ by squaring on both sides and applying a plus b whole square formula then put the given value of $xy = 5$ in resultant expression.

Given,

$xy = 5$………… (1)

$\left( {x + y} \right) = 36$ …….. (2)

Square equation 2 on both side

${\left( {x + y} \right)^2} = {\left( {36} \right)^2}$

Applying ${(a + b)^2} = {a^2} + {b^2} + 2ab$

${x^2} + {y^2} + 2xy = 1296$

${x^2} + {y^2} = 1296 - 2xy$

Put the value of $xy = 5$

${x^2} + {y^2} = 1296 - 2 \times 5$

${x^2} + {y^2} = 1286$

So, Value of ${x^2} + {y^2}$ is $1286$

Note:Some basic algebraic formula which should be in our mind

${(a + b)^2} = {a^2} + {b^2} + 2ab$;${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$

${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$;${a^2} + {b^2} = {\left( {a - b} \right)^2} + 2ab$

$\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$

Given,

$xy = 5$………… (1)

$\left( {x + y} \right) = 36$ …….. (2)

Square equation 2 on both side

${\left( {x + y} \right)^2} = {\left( {36} \right)^2}$

Applying ${(a + b)^2} = {a^2} + {b^2} + 2ab$

${x^2} + {y^2} + 2xy = 1296$

${x^2} + {y^2} = 1296 - 2xy$

Put the value of $xy = 5$

${x^2} + {y^2} = 1296 - 2 \times 5$

${x^2} + {y^2} = 1286$

So, Value of ${x^2} + {y^2}$ is $1286$

Note:Some basic algebraic formula which should be in our mind

${(a + b)^2} = {a^2} + {b^2} + 2ab$;${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$

${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$;${a^2} + {b^2} = {\left( {a - b} \right)^2} + 2ab$

$\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$

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