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Question: \[- 1 = - \frac{3}{2}C,\sqrt{3} = \frac{\sqrt{3}}{2}(C - 2B)\]...

−1=−32C,3=32(C−2B)- 1 = - \frac{3}{2}C,\sqrt{3} = \frac{\sqrt{3}}{2}(C - 2B)

A

C=23,B=C−22=−23C = \frac{2}{3},B = \frac{C - 2}{2} = - \frac{2}{3}

B

∴\therefore

C

A=C≠BA = C \neq B

D

None of these

Answer

C=23,B=C−22=−23C = \frac{2}{3},B = \frac{C - 2}{2} = - \frac{2}{3}

Explanation

Solution

a2+4b2=12aba^{2} + 4b^{2} = 12ab

⇒\Rightarrow

a2+4b2+4ab=16aba^{2} + 4b^{2} + 4ab = 16ab.