Question:
If $(1 - p)$ is a root of quadratic equation $x^2 + px + (1- p) = 0$, then its roots are
If $(1 - p)$ is a root of quadratic equation $x^2 + px + (1- p) = 0$, then its roots are
Updated On: Jul 28, 2022
- 0, 1
- -1, 2
- 0, -1
- -1, 1
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The Correct Option is C
Solution and Explanation
$\left(1-p\right)^{2}+p\left(1-p\right)+\left(1-p\right) = 0\quad$ (since $\left(1 - p\right)$ is a root of the equation $x^{2} + px + \left(1 - p\right) = 0$)
$?\quad \left(1- p\right)\left(1- p + p + 1\right) = 0$
$?\quad 2\left(1- p\right) = 0? \left(1 - p\right) = 0 ? p = 1$
sum of root is $a + ??= -p$ and product $a??= 1- p = 0\quad$ (where $??= 1 - p = 0$)
$?\quad a + 0 = -1 \quad? a = -1?\quad$ Roots are $0, -1$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
