In this question I encountered the quadratic equation Q1^2 + Q2^2 – 6(Q1)(Q2) = 0
How can I find the ration of Q1/Q2 ? How is the answer coming C please help .
Best Answer
Divide your equation by $Q2^2$. You will have a quadratic in $\frac {Q1}{Q2}$ which completing the square or the quadratic formula will solve.
Because of the symmetry of the problem some ratio for $\frac {Q1}{Q2}$ and its inverse must both be solutions because you can rename the two spheres and get the reciprocal ratio. Only choice c gives two choices, so it has to be right. You can verify $-(3+\sqrt 8)(-3+\sqrt 8)=1$ as a check.
Apparently, the problem is not that easy after all. What your teacher believes is the solution is obviously wrong. After all, she suggests that $\frac12$ is one of the solutions, but
$$ (\tfrac12)^2-(k+1)\tfrac12+k+1\ne 0$$
If $a=0$, then you have a straight line. Then $$f(1)=\frac{f(-1)+f(3)}2>\frac{-4+5}2=\frac 12>0$$
Then for a parabola with $a>0$ you have $$f(\frac{x+y}2)<\frac{f(x)+f(y)}2$$ and for $a<0$ you have $$f(\frac{x+y}2)>\frac{f(x)+f(y)}2$$
Since $$\frac{f(-1)+f(3)}2>0$$ and $$f(1)<0$$ then $$f(\frac{-1+3}2)<\frac{f(-1)+f(3)}2$$
Best Answer
Divide your equation by $Q2^2$. You will have a quadratic in $\frac {Q1}{Q2}$ which completing the square or the quadratic formula will solve.
Because of the symmetry of the problem some ratio for $\frac {Q1}{Q2}$ and its inverse must both be solutions because you can rename the two spheres and get the reciprocal ratio. Only choice c gives two choices, so it has to be right. You can verify $-(3+\sqrt 8)(-3+\sqrt 8)=1$ as a check.