Question

if a=1/2+root3 and b=1/2-root3 then find a square +b square -14ab.

Answer 1

the answer of the given problem is 45

Answer 2

The data given in the question is a=$\frac{1}{2} +\sqrt{3}$

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

we have to find $a^{2} + b^{2} -14ab$

$(\frac{1}{2} +\sqrt{3} )^{2} +(\frac{1}{2} -\sqrt{3})^{2} -14(\frac{1}{2} +\sqrt{3}) (\frac{1}{2} -\sqrt{3})$

by solving this we get

$(\frac{1+2\sqrt{3} }{2} )^{2} +(\frac{1-2\sqrt{3} }{2}) ^{2} -14(\frac{1+2\sqrt{3} }{2}) (\frac{1-2\sqrt{3} }{2} )$

By proceeding further with calculation we get

$\frac{1+12+4\sqrt{3} +1+12-4\sqrt{3} }{4} -- 14 \frac{(1-12)}{4}$

=$\frac{26}{4} -14(\frac{-11}{4} )$

=$\frac{26}{4} +\frac{154}{4}$

$\frac{180}{4}$

=45.

Therefore ,the final answer 45.

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