Question

Pls answer fast and correct no nonsense ​

Answer 1

✧ Question:

☞ If $\sf{a = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}}$ , $\sf{b = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}}$ , then the value of $\mathtt{(a + b)^{2}}$

✧ To Find:

☞ The value of $\mathtt{(a + b)^{2}}$

✧ Given:

☞ The value of :-

• $\sf{a = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}}$

• $\sf{b = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}}$

✧ We Know:

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

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

• $\it{a^{2} - b^{2} = (a + b)(a - b)}$

• $\it{a \times a = a^{2}}$

✧ Concept:

☞ By putting the value of a and b in the equation, we can find the value of $\mathtt{(a + b)^{2}}$.

✧ Solution:

☞ Given Equation:

$\mathtt{(a + b)^{2}}$

➜ Putting the value of a and b in the equation,

• $\sf{a = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}}$
• $\sf{b = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}}$

➜ we get,

$\mathtt{\Rightarrow \left(\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} + \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\right)^{2}}$

➝ Using the identity:

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

➝ we get,

$\mathtt{\Rightarrow \left(\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\right)^{2} + 2 \times \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \left(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\right)^{2}}$

$\mathtt{\Rightarrow \left(\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\right)^{2} + 2 \times \cancel{\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}} \times \cancel{\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}} + \left(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\right)^{2}}$

$\mathtt{\Rightarrow \left(\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\right)^{2} + 2 + \left(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\right)^{2}}$

$\mathtt{\Rightarrow \dfrac{\left(\sqrt{3} - \sqrt{2}\right)^{2}}{\left(\sqrt{3} + \sqrt{2}\right)^{2}} + 2 + \dfrac{\left(\sqrt{3} + \sqrt{2}\right)^{2}}{\left(\sqrt{3} - \sqrt{2}\right)^{2}}}$

➝ Using the identity:

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

and

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

➝ we get,

$\mathtt{\Rightarrow \dfrac{(\sqrt{3})^{2} - 2 \times \sqrt{3} \times \sqrt{2} + (\sqrt{2})^{2}}{(\sqrt{3})^{2} + 2 \times \sqrt{3} \times \sqrt{2} + (\sqrt{2})^{2}} + 2 + \dfrac{(\sqrt{3})^{2} + 2 \times \sqrt{3} \times \sqrt{2} (\sqrt{2})^{2}}{(\sqrt{3})^{2} - 2 \times \sqrt{3} \times \sqrt{2} + (\sqrt{2})^{2}}}$

$\mathtt{\Rightarrow \dfrac{5 - 2\sqrt{6}}{5 + 2\sqrt{6}} + 2 + \dfrac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}}$

LCM of the denominators is $(5 + 2\sqrt{6})(5 - 2\sqrt{6})$

$\mathtt{\Rightarrow \dfrac{25 - 10\sqrt{6} + 24)(5 + 2\sqrt{6}) + (5 + 2\sqrt{6})(5 + 2\sqrt{6})}{(5 + 2\sqrt{6})(5 - 2\sqrt{6})}}$

➝ Using the identity:

$\it{a \times a = a^{2}}$

➝ we get,

$\mathtt{\Rightarrow \dfrac{(5 - 2\sqrt{6})^{2} + 2(5 - 2\sqrt{6})(5 + 2\sqrt{6}) + (5 + \sqrt{6})^{2}}{(5 + 2\sqrt{6})(5 - 2\sqrt{6})}}$

➝ Using the identity:

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

and

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

➝ we get,

$\mathtt{\Rightarrow \dfrac{25 - 20\sqrt{6} + 24 + 2 \times 5^{2} - (2\sqrt{6})^{2} + 25 + 20\sqrt{6} + 24}{5^{2} - (2\sqrt{6})^{2}}}$

$\mathtt{\Rightarrow \dfrac{25 - \cancel{20\sqrt{6}} + 24 + 2 \times 25 - 2 \times (25 - 24) + 25 + \cancel{20\sqrt{6}} + 24}{25 - 24}}$

$\mathtt{\Rightarrow 25 + 24 + 2 + 25 + 24}$

$\mathtt{\Rightarrow 100}$

Hence, the value of $\mathtt{(a + b)^{2}}$ is 100.

Extra Information:

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

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

• $\bigg(x + \dfrac{1}{x}\bigg)^{2} = x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} + 2$

• $\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} - 2$

Answer 2

• thank ♥ the answer if u like it and it is helpful for you.

$\huge\mathtt\pink{To\:\:Find\::-}$

✒the value of ( a + b ) ²

➡$\large\mathtt\red{A}$$\large\mathtt\blue{N}$$\large\mathtt\green{S}$$\large\mathtt\pink{W}$$\large\mathtt{E}$$\large\mathtt\purple{R}$$\large\mathtt\orange{:-}$

✒a =

$<strong>a</strong> = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$= \frac{( \sqrt{3} - \sqrt{2} )( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} }$

$= \frac{( \sqrt{3} - \sqrt{2} ) {}^{2} }{3 - 2}$

$= (( \sqrt{3}) {}^{2} + ( \sqrt{2}) {}^{2} + - 2( \sqrt{3})( \sqrt{2})$

$= 3 + 2 - 2 \sqrt{6}$

$= 5 - 2 \sqrt{6}$

✒b =

$<strong>b</strong> = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$= \frac{( \sqrt{3 } + \sqrt{2} )( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} }$

$= \frac{( \sqrt{3} + \sqrt{2}) {}^{2} }{3 - 2}$

$= (( \sqrt{3} ) {}^{2} + \: \: ( \sqrt{2}) {}^{2} + 2( \sqrt{3} )( \sqrt{2})$

$= 3 + 2 + 3 \sqrt{6}$

$= 5 + 2 \sqrt{6}$

✒NOW,

$a { \: }^{2} + {b \: }^{2}$

• we know that

$(x + y) {}^{2} = x {}^{2} + y {}^{2} + 2xy$

$= > x {}^{2} + y {}^{2} = (x + y) {}^{2} - 2xy$

• if we take X = a and y = b then

${a}^{2} + {b}^{2}$

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

$= (5 + 2 \sqrt{6} + 5 - 2 \sqrt{6}) {}^{2}$

$- 2((5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ))$

$= (10) {}^{2} - 2(25 - 24)$

• Here we use identity

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

$= 100 - 2$

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$= 98$

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hope it help uu

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