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

If the roots of the equation

x^{4} + p x^{3} + q x^{2} + r x + s = 0

are in arithmetical progression, show that

p^{3} - 4 p q + 8 r = 0

and if they are in geometrical progression, show that

p^{2} s = r^{2}

.

Accepted Answer

Given equation is $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ .

Suppose its roots are in $A . P .$ , they are $α - 3 β, α - β, α + β, α + 3 β$ .

Then sum of roots, $4 α = - p . . . (i)$

Sum of products of roots taken two at a time, $6 α^{2} - 10 β^{2} = q . . . (ii)$

and $4 α^{2} - 20 β^{2} = - r . . . (iii)$

From $(i)$ and $(ii)$ , we can write

$α = - \frac{p}{4}$ and $6 {(- \frac{p}{4})}^{2} - 10 β^{2} = q$

or $- 10 β^{2} = q - \frac{3}{8} p^{2}$

Putting this value of $β^{2}$ and that of $α$ in $(iii),$ we get

$4 {(\frac{- p}{4})}^{2} + 2 (\frac{- p}{4}) (q - \frac{3}{8} p^{2}) = - r$

$\Rightarrow \frac{- p^{2}}{16} - \frac{p}{2} (q - \frac{3}{8} p^{2}) = - r$

$\Rightarrow - p^{3} - 8 p q + 3 p^{3} = - 16 r$

$\Rightarrow 2 p^{3} - 8 p q + 16 r = 0$

$\Rightarrow p^{3} - 4 p q + 8 r = 0$

Part- $II$ of question

Given equation is $x^{4} + p x^{3} + q x^{2} + r x + s = 0$

Let the roots in $G . P .$ be $\frac{α}{β^{3}}, \frac{α}{β}, α β, α β^{2}$
The sum of roots is $α (\frac{1}{β^{3}} + \frac{1}{β} + β + β^{3}) = - p . . . (i)$
Sum of products of roots taken three at a time is

$\frac{α^{3}}{β^{3}} + \frac{α^{3}}{β} + α^{2} β + α^{3} β^{3} = - r . . . (ii)$

And product of roots $α^{4} = s . . . (iii)$

From $(i)$ and $(ii),$ we get

$α^{2} = \frac{- p}{- r} = \frac{p}{r}$

Putting this value of $α^{2}$ in $(iii),$ we get, $\frac{p^{2}}{r^{2}} = s$

or $p^{2} = r^{2} s$ is the required condition.

If the roots of the equation $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ are in arithmetical progression, show that $p^{3} - 4 p q + 8 r = 0$ and if they are in geometrical progression, show that $p^{2} s = r^{2}$ .

Important Questions on Theory of Equations

Learn and Prepare for any exam you want !

If the roots of the equation x4+px3+qx2+rx+s=0 are in arithmetical progression, show that p3-4pq+8r=0 and if they are in geometrical progression, show that p2s=r2.

Important Questions on Theory of Equations

Learn and Prepare for any exam you want !

If the roots of the equation $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ are in arithmetical progression, show that $p^{3} - 4 p q + 8 r = 0$ and if they are in geometrical progression, show that $p^{2} s = r^{2}$ .