Question

if the roots of 10x^3 -cx^2 -54x -27 =0 are in harmonic progression .then find c and all the roots

Answer 1

The given equation is $10x^3-cx^2-54x-27=0$

Let the roots be p, q, r

So

p + q + r = $\frac{c}{10}$ .......(1)

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

pqr= $\frac{27}{10}$ ........(3)

Now as these roots are in HP

So $\frac{1}{p},\frac{1}{q}, \frac{1}{r}$ are in AP

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

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

2pr=pq + qr

2pr = $\frac{-54}{10}$ -pr (From (2))

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

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

But pqr = $\frac{27}{10}$

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

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

Substituting the values of pr & q in (4) we get

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

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

Substituting in (1)

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

c= 9

The equation becomes

$10x^3-9x^2-54x-27=0$

Factoring we get

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

The roots are 3, -3/2, -3/5