Question

evaluate using identities.​

Answer 1

Algebraic Identities

The following standard algebraic identity will be used to solve the problem:

$\[\boxed{a^2 + b^2 = a^2 + b^2 + 2ab}\]$

We have been given that, $a + b = 7$ and $ab = 10$. With this information, we have been asked to find out the value of $a^2 + b^2$.

We know that,

$\[\implies \boxed{a^2 + b^2 = a^2 + b^2 + 2ab}\]$

By substituting the given values in the above identity, we get:

$\implies a^2 + b^2 = a^2 + b^2 + 2(10) \\ \\ \implies (a + b)^2 = a² + b² + 20 \\ \\ \implies {7}^{2} = a^2 + b^2 + 20 \\ \\ \implies 49 = a^2 + b^2 + 20 \\ \\ \implies 49 - 20 = {a}^{2} + {b}^{2} \\ \\ \implies \boxed{{a}^{2} + {b}^{2} = 29}$

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MORE TO KNOW

$\boxed{\begin{array}{l}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\\frak{1.}\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\\frak{2.}\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\\frak{3.}\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\\frak{4.}\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\\frak{5.}\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\\frak{6.}\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\\frak{7.}\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\\frak{8.}\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}$

Answer 2

$\underline{ \pink{ \rm GIVEN \: \: DATA\: \: : }}$

$\rm a + b = 7$

$\rm ab = 10$

$\underline{ \purple{ \rm TO \: \: FIND \: \: : }}$

$\rm the \: \: \: value \: \: of \: \: \boxed{ \rm {a}^{2} + {b}^{2} }$

$✯ \: \underline{ \red{ \rm SOLUTION \: \: : }}$

By using the identity,

$\boxed{ ⍟ \: \: \color{blue}{ \boxed{\boxed{ \color{black} \bf {a}^{2} + {b}^{2} = {a}^{2} + {b}^{2} + 2ab}}}}$

so by applying the values,

$\rm➠ \: \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2(10)$

$\rm➠ \: \: {(7)}^{2} = {a}^{2} + {b}^{2} + 20$

$\rm➠ \: \: {a}^{2} + {b}^{2} = 49 - 20$

$\color{cyan}{ \boxed{ \boxed{ \red\ast \: \: \color{brown} \boxed{ \boxed{ \green{ \rm {a}^{2} + {b}^{2} = 29}}}}}}$