Name: Transformation of Roots of a Quadratic Equation
Uploaded: 2019-07-19
Duration: 11 min 45 s
Description: Here an equation and its roots are given, and we have to find new quadratic equations with new roots, which are simple multiples or factors of the original r...

Question

Find the condition that

x^{3} - p x^{2} + q x - r = 0

may have,
(1) two roots equal but of opposite sign.
(2) the roots in geometrical progression.

Accepted Answer

(1) Given equation is $x^{3} - p x^{2} + q x - r = 0$

As per the given conditions, suppose its roots are $α, - α, β$

Then sum of roots $β = p . . . (i)$ and sum of the products of roots taken two at a time is $- α^{2} = q . . . (ii)$ and product of roots $- α^{2} β = r . . . (iii)$

The elimination of $α, β$ from $(i), (ii), (iii)$ is the required condition.

From $(ii)$ and $(iii)$

$β = \frac{+ r}{q} ∴ p = \frac{+ r}{q}$

or $p q = r$ is the required condition.

$(2) x^{3} - p x^{2} + q x - r = 0$

Let the roots in $G . P .$ be $\frac{α}{β}, α, α β$ , then $α^{3} = r . . . (i)$

Sum of roots is $α (\frac{1}{β} + 1 + β) = p . . . (ii)$
Sum of products of roots taken two at time is $\frac{α^{2}}{β} + α^{2} + α^{2} β = q$

or $α^{2} (\frac{1}{β} + 1 + β) = q . . . (iii)$

The elimination of $α, β$ from the above equations will give us the required condition.

From $(ii)$ and $(iii)$ , we get $α = \frac{q}{p}$

then ${(\frac{q}{p})}^{3} = r$

or $q^{3} = p^{3} r$ is the required condition.

Find the condition that $x^{3} - p x^{2} + q x - r = 0$ may have,
(1) two roots equal but of opposite sign.
(2) the roots in geometrical progression.

Important Questions on Theory of Equations

Learn and Prepare for any exam you want !

Find the condition that x3-px2+qx-r=0 may have, (1) two roots equal but of opposite sign. (2) the roots in geometrical progression.

Important Questions on Theory of Equations

Learn and Prepare for any exam you want !

Find the condition that $x^{3} - p x^{2} + q x - r = 0$ may have,
(1) two roots equal but of opposite sign.
(2) the roots in geometrical progression.