Question

if
$a = 1 \div 7 - 4 \sqrt{3}$
$b \: = 1 \div 7 + 4 \sqrt{3}$
evaluate - a²+b²

​

Answer 1

Solution:-

Given

$\sf \to \: a = \dfrac{1}{7 - 4 \sqrt{3} } \: \: and \: \: b = \dfrac{1}{7 + 4 \sqrt{3} }$

To find the value of

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

Put the value on formula

$\sf \to \: \bigg( \dfrac{1}{7 - 4 \sqrt{3} } \bigg)^{2} + \bigg( \dfrac{1}{7 + 4 \sqrt{3} } \bigg)^{2}$

$\sf \to \: \dfrac{1}{(7 - 4 \sqrt{3} ) {}^{2} } + \dfrac{1}{(7 + 4 \sqrt{3} ) {}^{2} }$

$\sf \to \: \dfrac{(7 + 4 \sqrt{3} ) {}^{2} + (7 - 4 \sqrt{3} ) {}^{2} }{(7 - 4 \sqrt{3} ) {}^{2} (7 + 4 \sqrt{3} ) {}^{2} }$

$\sf \to \: \dfrac{(7 + 4 \sqrt{3} ) {}^{2} + (7 - 4 \sqrt{3} ) {}^{2} }{(7 - 4 \sqrt{3} ) {}^{} \times (7 - 4 \sqrt{3}) {}^{} \times (7 + 4 \sqrt{3} ) {}^{} \times (7 + 4 \sqrt{3} )}$

Using this identity

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

$\sf \to \: \dfrac{(7) {}^{2} + (4 \sqrt{3} ) {}^{2} - 2 \times 7 \times 4 \sqrt{3} + (7) {}^{2} + (4 \sqrt{3} ) {}^{2} + 2 \times 7 \times 4 \sqrt{3} }{(( {7})^{2} - (4 \sqrt{3} ) ^{2} ) \times (( {7})^{2} - (4 \sqrt{3} ) ^{2} ) }$

$\sf \to \: \dfrac{49 + 48 + 49 + 48}{(49 - 48)(49 - 48)}$

$\sf \to \: 98 + 96$

$\sf \to \: 194$

Answer is 194

Answer 2

We have,

$a = \frac{1}{7 - 4 \sqrt{3} }$

On rationalising the denominator,

$\implies a = 7 + 4 \sqrt{3}$.

Also, $b = \frac{1}{7 + 4 \sqrt{3} }$.

Similarly,

$\implies b = 7 - 4 \sqrt{3}$.

We know,

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

[ab = (7 + 4√3)(7 - 4√3) = 1]

$\therefore a^2 - b^2 = {[(7 + 4 \sqrt{3}) + (7 - 4 \sqrt{3}]}^{2} - 2$

$\implies a^2 - b^2 = (14)^2 - 2$

$\implies a^2 - b^2 = 196 - 2$

$\implies a^2 - b^2 = 194$. (Answer.)