Question

If roots of 10x^3-cx^2-54x-27=0 are in H.P.,find c.

Answer 1

The given equation is

$10x^3-cx^2-54x-27=0$

Let the roots be p, q and r.

So, $p+q+r = \frac{c}{10}$  (Equation 1)

$pq+qr+pr= \frac{-54}{10}$   (Equation 2)

$pqr = \frac{27}{10}$  (Equation 3)

Since, these roots are in H.P.

So, $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in A.P.

$\frac{2}{q}=\frac{1}{p}+\frac{1}{r}$

2pr = q(p+r) (Equation 4)

2pr = pq+qr

2pr = $\frac{-54}{10} - pr$

3pr = $\frac{-54}{10}$

pr = $\frac{-18}{10}$

Since, $pqr = \frac{27}{10}$

$\frac{-18}{10}q=\frac{27}{10}$

So, q=$\frac{-3}{2}$

Substituting the values of pr and q in 4, we get

$2 \times \frac{-18}{10} = \frac{-3}{2}(p+r)$

$p+r = \frac{24}{10}$

Substituting this in 1,

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

c = 9

Therefore, the equation becomes $10x^3-9x^2-54x-27=0$.

Factorizing we get,

(x-3)(2x+3)(5x+3)=0

The roots are 3, $\frac{-3}{2}$ and $\frac{-3}{5}$.