Question

Solve the equation,

27 x^{4} - 195 x^{3} + 494 x^{2} - 520 x + 192 = 0

, the roots being in geometrical progression.

Accepted Answer

Let us assume a polynomial equation is

$f (x) = a x^{n} + b x^{n - 1} + c x^{n - 2} + \dots + z$

Then, we can say the sum of all roots is $- \frac{b}{a}$ . If the degree of the polynomial is even, then the product of the roots is $\frac{z}{a}$ and if the degree of the polynomial is odd, then the product of the roots is $- \frac{z}{a}$ . The sum of the products of the roots taken two at a time is $\frac{c}{a}$ .

Given equation is, $27 x^{4} - 195 x^{3} + 494 x^{2} - 520 x + 192 = 0$

The degree of the polynomial is even. Then, the product of the roots is $\frac{192}{27} = \frac{64}{9}$ .

The sum of the products of the roots taken two at a time is $\frac{494}{27}$ ,

It is given the roots of the equation are in geometric progression. Let us assume the roots are $\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$ .

Then product of the roots,

$\frac{a}{r^{3}} \times \frac{a}{r} \times a r \times a r^{3} = \frac{64}{9}$

$\Rightarrow a^{4} = \frac{64}{9}$

$∴ a^{2} = \frac{8}{3}$

Sum of products of roots taken two at a time,

$\frac{a}{r^{3}} \times \frac{a}{r} + \frac{a}{r^{3}} \times a r + \frac{a}{r^{3}} \times a r^{3} + \frac{a}{r} \times a r + \frac{a}{r} \times a r^{3} + a r \times a r^{3} = \frac{494}{27}$

$\Rightarrow a^{2} (\frac{1}{r^{4}} + \frac{1}{r^{2}} + 2 + r^{2} + r^{4}) = \frac{494}{27}$

$\Rightarrow \frac{8}{3} \times (\frac{1}{r^{4}} + \frac{1}{r^{2}} + 2 \times r^{2} \times \frac{1}{r^{2}} + r^{2} + \frac{1}{r^{2}}) = \frac{494}{27}$

$\Rightarrow {(\frac{1}{r^{2}} + r^{2})}^{2} + (r^{2} + \frac{1}{r^{2}}) = \frac{247}{36}$

Now, we have put $r^{2} + \frac{1}{r^{2}} = p,$ to simply the above quadratic equation.

Now, the above equation reduces as,

$p^{2} + p - \frac{247}{36} = 0$

$\Rightarrow 36 p^{2} + 36 p - 247 = 0$

$\Rightarrow 36 p^{2} - 78 p + 114 p - 247 = 0$

$\Rightarrow (6 p - 13) (6 p + 19) = 0$

$\Rightarrow p = \frac{13}{6}$ or $p = \frac{- 19}{6}$

We cannot use $p = - \frac{19}{6}$ because $p$ is the sum of two square terms and the sum of square terms is always a positive number.

$∴ p = \frac{13}{6}$

$\Rightarrow r^{2} + \frac{1}{r^{2}} = \frac{13}{6}$

$\Rightarrow r^{4} + 1 = \frac{13}{6} \times r^{2}$

$\Rightarrow 6 r^{4} - 13 r^{2} + 6 = 0$

$\Rightarrow 6 r^{4} - 4 r^{2} - 9 r^{2} + 6 = 0$

$\Rightarrow (3 r^{2} - 2) (2 r^{2} - 3) = 0$

$\Rightarrow r^{2} = \frac{3}{2}$ or $r^{2} = \frac{2}{3}$

Here, we can choose any value of $r^{2}$ ,

For now, we have choose $r^{2} = \frac{3}{2}$ .

$∴ a^{2} r^{2} = \frac{8}{3} \times \frac{3}{2} = 4$

$∴ a r = 2$

Thus, the roots of the equation are $3, 2, \frac{4}{3}, \frac{8}{9}$ .

H S Hall and S R Knight Solutions for Chapter: Theory of Equations, Exercise 1: EXAMPLES. XXXV. a.

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Theory of Equations, Exercise 1: EXAMPLES. XXXV. a.

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