Question

(a-b) = 7 and ab=9, then (a+b)^2 =?​

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given that
• ( a - b ) = 7
• ab = 9

To Find:

• To Find the value of ( a + b )²

Solution:

We have been given that

$\boxed{\sf{ a - b = 7}}$

$\boxed{\sf{ ab = 9}}$

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Squaring both sides in First Equation

$\implies \sf{ ( a - b )^2 = (7)^2}$

$\implies \sf{ a^2 + b^2 - 2ab = 49}$

$\implies \sf{ a^2 + b^2 - 2(9) = 49}$

$\implies \sf{ a^2 + b^2 - 18 = 49}$

$\implies \sf{ a^2 + b^2 = 49 + 18}$

$\implies \boxed{\sf{ a^2 + b^2 = 67}}$

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$\large\underline{\mathfrak\orange{Finding \: the \: value \: of \: given \: expression}}$

$\implies \sf{( a + b )^2}$

$\implies \sf{ a^2 + b^2 + 2ab}$

$\implies \sf{ 67 + 2 \: (9)}$

$\implies \sf{ 67 + 18}$

$\implies \sf{ 85}$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\large\boxed{\sf{\red{ ( a + b )^2 = 85}}}$

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$\underline{\large\sf{\purple{Some \: Basic \: Algebric \: Identities }} }$

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

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

$\sf{(a+b)^3 = a^3 + b^3 + 3ab(a+b)}$

$\sf{(a-b)^3 = a^3 - b^3 - 3ab(a-b)}$

$\sf{a^3 + b^3 = (a+b)(a^2 + b^2 - ab)}$

$\sf{a^3 - b^3 = (a-b)(a^2 + b^2 + ab)}$

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Answer 2

Answer:

❤️HELLO❤️MATE❤️

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Given:-

• $(a - b) = 7$
• $ab = 9$

To find:-

• ${(a + b)}^{2}$

Solution:-

We have ,

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

putting the given value,

$= > {7}^{2} = {a}^{2} + {b}^{2} - 2.9$

$= > {a}^{2} + {b}^{2} = 49 + 18$

$= > {a}^{2} + {b}^{2} = 67$

Now,

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

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 67 + 18$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 85$

Hence,the value of ${a}^{2}+{b}^{2}$ is 85 .

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