If alpha and beta are the roots of the equati | vedclass

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

Question

(C) Given the quadratic equation: ${x^2} - a(x + 1) - b = 0$Expanding the equation: ${x^2} - ax - a - b = 0$Comparing this with the standard form ${x^

Vedclass · Accepted Answer

$1 - b$

Vedclass · Answer

(C) Given the quadratic equation: ${x^2} - a(x + 1) - b = 0$Expanding the equation: ${x^2} - ax - a - b = 0$Comparing this with the standard form ${x^2} - (	ext{sum of roots})x + (	ext{product of roots}) = 0$,we get:Sum of roots: $\alpha + \beta = a$Product of roots: $\alpha \beta = -(a + b)$We need to evaluate the expression: $(\alpha + 1)(\beta + 1)$Expanding the expression: $(\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1$Substituting the values of $\alpha + \beta$ and $\alpha \beta$:$= -(a + b) + a + 1$$= -a - b + a + 1$$= 1 - b$

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - a(x + 1) - b = 0$,then $(\alpha + 1)(\beta + 1) = $

Similar Questions

The number of real roots of the equation $5 + |2^x - 1| = 2^x(2^x - 2)$ is

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 3 = 0$,then the equation whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{784} x + 1234 = 1486$
$II.$ $\sqrt{1089} y + 2081 = 2345$

The number of real roots of the equation $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$,where $P$ and $Q$ are non-zero real numbers,is

If one root of the equation ${x^2} + px + q = 0$ is $2 + \sqrt{3}$,then the values of $p$ and $q$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module