The sum of two roots of the equations $x^{4} - 8 x^{3} + 21 x^{2} - 20 x + 5 = 0$ is $4$ ; explain why on attempting to solve the equation from the knowledge of this fact the method fails.

Open in App

Solution

$x^{4} - 8 x^{3} + 21 x^{2} - 20 x + 5 = (x^{2} - 5 x + 5) (x^{2} - 3 x + 1)$ ;
As some of two roots is $4$ , put $x = 4 - y$
The expressions $x^{2} - 5 x + 5$ and $x^{2} - 3 x + 1$ become $y^{2} - 3 y + 1$ and $y^{2} - 5 y + 5$ respectively.
Then, we get $(y^{2} - 3 y + 1)$ $(y^{2} - 5 y + 5) = y^{4} - 8 x^{3} + 21 x^{2} - 20 x + 5$ which is the original equation.
So, this method fails as we get the same equation after substituting the relation between the roots.

Suggest Corrections

Similar questions

x^{4} - 8 x^{3} + 21 x^{2} - 20 x + 5 = 0

5

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

x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 20 = 0

3 + i

x^{4} + 8 x^{3} + 9 x^{2} - 8 x - 10 = 0

x^{4} - 10 x^{2} - 20 x - 16 = 0

Join BYJU'S Learning Program