Question

If a =
1/2-√3

and b = 1/2+√3

then find a2+ab+b2​

Answer 1

Answer:

$a = (1 \div 2 - \sqrt{3} ) \\ b = (1 \div 2 + \sqrt{3} ) \\ \\ {a}^{2} + ab + {b}^{2} \\ = {a}^{2} + 2ab + {b}^{2} - ab \\ = {(a + b)}^{2} - ab \\ = {(1 \div 2 - \sqrt{3} + 1 \div 2 + \sqrt{3} )}^{2} - ((1 \div 2 - \sqrt{3} )(1 \div 2 + \sqrt{3} )) \\ = {1}^{2} - (1 \div 4 - 3) \\ = 1 - ( - 11 \div 4) \\ = 1 + 11 \div 4 \\ = 15 \div 4 \\ = 3 \times \frac{3}{4}$

Answer 2

given

$a = \frac{1}{2 - \sqrt{3} } \\ \\ b = \frac{1}{2 + \sqrt{3} }$

-----(1)

to find

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

----------(2)

solution=>

EQ 1 value put eq 2

$=( \frac{1}{2 - \sqrt{3} }) {}^{2} + \frac{1}{(2 - \sqrt{3})(2 + \sqrt{3}) } + ( \frac{1}{2 + \sqrt{3} }) {}^{2} \\ \\ = \frac{1}{(2 - \sqrt{3}) {}^{2} } + \frac{1}{2 {}^{2} - ( \sqrt{3} ) {}^{2} } + \frac{1}{(2 + \sqrt{3}) {}^{2} } \\ \\ = \frac{1}{4 + 3 - 4 \sqrt{3} } + 1 + \frac{1}{4 + 3 + 4 \sqrt{3} } \\ \\ = \frac{1}{7 - 4 \sqrt{3} } + \frac{1}{7 + 4 \sqrt{3} } + 1 \\ \\ = \frac{14}{49 - 48} + 1 \\ \\ = 14 + 1 = 15$

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formula uses

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

I hope it helps you