Question

The condition that the roots of x3-bx2+cx-d=0 are in G.P. is

Answer 1

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Answer 2

The condition that the roots of $x^{3} -bx^{2} +cx-d=0$ are in G.P. is $c^{3} =b^{3} d$.

Step-by-step explanation:

Given:

The roots of $x^{3} -bx^{2} +cx-d=0$ are in G.P.

To Find:

The condition that the roots of $x^{3} -bx^{2} +cx-d=0$ are in G.P.

Formula Used:

If $\alpha ,\beta\ and \ \gamma$ are the roots of equation $ax^{3} +bx^{2} +cx+d=0$.

Then  $\alpha \beta\ \gamma=-\frac{d}{a}$                   ------------- formula no.01.

Solution:

As given-the roots of $x^{3} -bx^{2} +cx-d=0$ are in G.P.

Let  $\frac{p}{q} , p\ and\ pq$ which are in G.P. are roots of equation $x^{3} -bx^{2} +cx-d=0$.

Similarly like formula no.01.

Products of roots $\frac{p}{q} \times p\ \times\ pq= -\frac{(-d)}{1}$

                             $\frac{p}{q} \times p\ \times\ pq= d$

                              $p^{3} =d$     ---------------------- equation no.01.

If p is the root of equation  $x^{3} -bx^{2} +cx-d=0$.

Then  $x=p$ should satisfy the equation $x^{3} -bx^{2} +cx-d=0$.

It means  $p^{3} -bp^{2} +cp-d=0$

Putting value of $p^{3}$ from equation no.01.

               $d -bp^{2} +cp-d=0$

                $cp=bp^{2}$

               $c=bp$

Cubing both sides.

          $c^{3} =(bp)^{3}$

          $c^{3} =b^{3} p^{3}$

From equation no.01, putting $p^{3} =d$.

          $c^{3} =b^{3} d$

Thus,the condition that the roots of $x^{3} -bx^{2} +cx-d=0$ are in G.P. is $c^{3} =b^{3} d$.

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