First slide

Theory of equations

Question

If $A, G$ and Hare respectively the A.M., G.M. and H.M. of three positive numbers $a, b$ and $c$ , then the equation whose roots are $a, b, c$ is

Moderate

A

$x^{3} - 3 A x^{2} + \frac{3 G^{3}}{H} x - G^{3} = 0$
B

$x^{3} + 3 A x^{2} + \frac{3 G^{3}}{H} x - G^{3} = 0$
C

$x^{3} + A x^{2} + \frac{G^{3}}{H} - G^{3} = 0$
D

$x^{3} - 3 A x^{2} - \frac{3 G^{3}}{H} x - G^{3} = 0$

Solution

By definition, we have
$\begin{aligned} A = \frac{a + b + c}{3} \Rightarrow a + b + c = 3 A \\ G = (a b c)^{1 / 3} \Rightarrow a b c = G^{3} \end{aligned}$
and, $\begin{array}{l} \frac{1}{H} = \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \\ \frac{3}{H} = \frac{b c + c a + a b}{a b c} \\ \frac{3}{H} = \frac{a b + b c + c a}{G^{3}} \Rightarrow a b + b c + c a = \frac{3 G^{3}}{H} \end{array}$
The equation whose roots are a, b and c is
$\begin{array}{l} x^{3} - S_{1} x^{2} + S_{2} x - S_{3} = 0 \\ x^{3} - (a + b + c) x^{2} + (a b + b c + c a) x - a b c = 0 \\ x^{3} - 3 A x^{2} + \frac{3 G^{3}}{H} x - G^{3} = 0 [Using (i), (ii) and (iii)] \end{array}$

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