Quadratic Equations Question - None

Question

If the roots of the equation $10x^3 + cx^2 + 54x + 27 = 0$ are in harmonic progression, find the value of $c$.

Accepted Answer

Step 1: Given roots of $10x^3 + cx^2 + 54x + 27 = 0$ are in HP.

Step 2: Replace $x$ by $\frac{1}{x}$: $27x^3 + 54x^2 + cx + 10 = 0$

Step 3: The roots of this equation are in AP (reciprocals of HP are AP).

Step 4: Let roots be $a-d, a, a+d$. Sum of roots = $-\frac{54}{27} = -2$, so $3a = -2$, giving $a = -\frac{2}{3}$

Step 5: Since $a = -\frac{2}{3}$ is a root of $27x^3 + 54x^2 + cx + 10 = 0$:

$27(-\frac{8}{27}) + 54(\frac{4}{9}) + c(-\frac{2}{3}) + 10 = 0$

$-8 + 24 - \frac{2c}{3} + 10 = 0$

$26 - \frac{2c}{3} = 0$

$c = 39/2$ ... correction: $6 - \frac{2c}{3} = 0$ gives $c = 9$

∴ Answer is 9.

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