- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
medium
The roots of the equation ${x^4} - 4{x^3} + 6{x^2} - 4x + 1 = 0$ are
A
$1, 1, 1, 1$
B
$2, 2, 2, 2$
C
$3, 1, 3, 1$
D
$1, 2, 1, 2$
Solution
(a) Let $f(x) = {x^4} – 4{x^3} + 6{x^2} – 4x + 1 = 0$
==> $(x – 1)({x^3} – 3{x^2} + 3x – 1) = 0$
==> $(x – 1){(x – 1)^3} = 0$
==> $x = 1,\,1,\,1,\,1$
Standard 11
Mathematics