If #x=a+bsqrt3#, where #a# and #b# are integers, and #x^2=97+56sqrt3#, what is the value of #ab#?

1 Answer
Jim G.
Apr 11, 2017

#ab=28#

Explanation:

Calculate #x^2" where " x=a+bsqrt3#

#rArrx^2=(a+bsqrt3)(a+bsqrt3)#

Expand brackets using FOIL method.

#rArrx^2=a^2+ab sqrt3+ab sqrt3+3b^2#

#color(white)(rArrx^2)=a^2+2ab sqrt3+3b^2#

#"Now "color(red)(a^2+3b^2)color(magenta)(+2ab sqrt3)=color(red)(97)color(magenta)(+56 sqrt3)=x^2#

#"Comparing the 2 sides of the identity"#

For the 2 sides to be equal then like terms have to #color(blue)"match "#Note we are not solving an equation here.

#rArrcolor(red)(a^2+3b^2)=color(red)(97)#

#"and " color(magenta)(2ab sqrt3=56 sqrt3)#

#rArr2abcancel(sqrt3)=56cancel(sqrt3)#

divide both sides by 2

#(cancel(2) ab)/cancel(2)=56/2#

#rArrab=28#

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