Solve the equations:
$x^{4} - 2 x^{3} - 21 x^{2} + 22 x + 40 = 0$ , the roots being in arithmetical progression.

Open in App

Solution

$x^{4} - 2 x^{3} - 21 x^{2} + 22 x + 40 = 0$

Let the roots be $a - 3 d, a - d, a + d, a + 3 d$

$S_{1} = a - 3 d + a - d + a + d + a + 3 d = - \frac{- 2}{1} 4 a = 2 a = \frac{1}{2} S_{4} = (a - 3 d) (a + 3 d) (a - d) (a + d) = \frac{40}{1} (a^{2} - 9 d^{2}) (a^{2} - d^{2}) = 40 (\frac{1}{4} - 9 d^{2}) (\frac{1}{4} - d^{2}) = 40 (1 - 36 d^{2}) (1 - 4 d^{2}) = 640 1 - 4 d^{2} - 36 d^{2} + 144 d^{4} - 640 = 0 144 d^{4} - 40 d^{2} - 639 = 0 144 d^{4} + 284 d^{2} - 324 d^{2} - 639 = 0 (4 d^{2} - 9) (36 d^{2} + 71) = 0 4 d^{2} - 9 = 0 \Rightarrow d = \pm \frac{3}{2}$

So the roots are $- 4, - 1, 2, 5$

Suggest Corrections

Similar questions

32 x^{3} - 48 x^{2} + 22 x - 3 = 0

4 x^{3} - 24 x^{2} + 23 x + 18 = 0

x^{4} + 2 x^{3} - 16 x^{2} - 22 x + 7 = 0

2 + \sqrt{3}

40 x^{4} - 22 x^{3} - 21 x^{2} + 2 x + 1 = 0

2 x^{3} - x^{2} - 22 x - 24 = 0

3 : 4

Join BYJU'S Learning Program