If (1 - p) is a root of the quadratic equatio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the quadratic equation $x^2 + px + (1 - p) = 0$ $(i)$.Since $(1 - p)$ is a root,it must satisfy the equation:$(1 - p)^2 + p(1 - p) + (1 - p)

Vedclass · Accepted Answer

$0, -1$

Vedclass · Answer

(C) Given the quadratic equation $x^2 + px + (1 - p) = 0$ $(i)$.Since $(1 - p)$ is a root,it must satisfy the equation:$(1 - p)^2 + p(1 - p) + (1 - p) = 0$Factor out $(1 - p)$:$(1 - p) [(1 - p) + p + 1] = 0$$(1 - p) [2] = 0$This implies $1 - p = 0$,so $p = 1$.Substitute $p = 1$ into the original equation $(i)$:$x^2 + (1)x + (1 - 1) = 0$$x^2 + x = 0$$x(x + 1) = 0$Thus,the roots are $x = 0$ and $x = -1$.

If $(1 - p)$ is a root of the quadratic equation $x^2 + px + (1 - p) = 0$,then its roots are:

Similar Questions

If $x+\frac{1}{x}=5,$ then $x^{6}+\frac{1}{x^{6}}$ is

The solution of the equation $x + \frac{1}{x} = 2$ is:

Sum of roots is $-1$ and sum of their reciprocals is $\frac{1}{6}$,then the equation is

The value of $k$ for which the quadratic equation $(k + 11)x^2 - (k + 3)x + 1 = 0$ has real and equal roots is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module