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.

H S Hall and S R Knight Solutions for Chapter: Theory of Equations, Exercise 2: EXAMPLES. XXXV. b.

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Theory of Equations, Exercise 2: EXAMPLES. XXXV. b.

Questions from H S Hall and S R Knight Solutions for Chapter: Theory of Equations, Exercise 2: EXAMPLES. XXXV. b. with Hints & Solutions

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