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# Three equations with square variables

Let

a^2+b^2+ab=9

b^2+c^2+bc=16

a^2+c^2+ac=25

What is the value of ab + bc +ac?

Note by Sinuhé Ancelmo
3 years, 8 months ago

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Take a point $$O$$ from which three line segments $$OA, OB, OC$$, lying on the same plane, emanate such that their respective lengths are $$a, b, c$$ and the angle between any two of them is 120 degrees. Use Cosine rule and the mentioned equations to find $$AB, BC, CA$$ which will have values 3,4 and 5 units. Now this triangle $$ABC$$ is right angled( by the converse of Pythagoras' Theorem), so it's area will be 6 sq. units. Also, the area of the three triangles $$AOB, BOC, COA$$ can be found out by the sine rule (individually), and now equate it with the earlier found area. The answer comes out to be $$8 \sqrt{3}$$. [ the calculations are to be done by you]. · 3 years, 8 months ago

Is it possible to solve this algebraically? · 3 years, 8 months ago

Well, your idea is interesting, but are you sure about the value of ab + bc +ac? shouldn' it be 8*sqrt(3)? Could you chek it again, plz?? · 3 years, 8 months ago

Could you show any algebraic, only algebraic solution, friend? · 3 years, 8 months ago

Edited. · 3 years, 8 months ago

You can realize that - 8 * sqrt (3) is also a valid answer for ab + bc + ac, right? · 3 years, 8 months ago

what will be the algebric solution Help me · 2 years ago