If the roots of x^2 - bx + c = 0 are two cons | vedclass

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

Question

(A) Let the roots be $\alpha$ and $\alpha + 1$.According to the properties of quadratic equations,the sum of the roots is $\alpha + (\alpha + 1) = 2\a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the roots be $\alpha$ and $\alpha + 1$.According to the properties of quadratic equations,the sum of the roots is $\alpha + (\alpha + 1) = 2\alpha + 1 = b$.The product of the roots is $\alpha(\alpha + 1) = c$.We need to find the value of the discriminant $D = b^2 - 4c$.Substituting the expressions for $b$ and $c$:$b^2 - 4c = (2\alpha + 1)^2 - 4(\alpha(\alpha + 1))$Expanding the terms:$b^2 - 4c = (4\alpha^2 + 4\alpha + 1) - (4\alpha^2 + 4\alpha)$Simplifying the expression:$b^2 - 4c = 4\alpha^2 + 4\alpha + 1 - 4\alpha^2 - 4\alpha = 1$.Therefore,$b^2 - 4c = 1$.

If the roots of $x^2 - bx + c = 0$ are two consecutive integers,then $b^2 - 4c$ is

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 6x + \lambda = 0$ and $3\alpha + 2\beta = -20$,then $\lambda = $

With respect to the roots of $x^{2}-x-2=0$,we can say that

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

Solve the given two equations and select the correct answer from the given options.
$I.$ $5x^{2} + 2x - 3 = 0$
$II.$ $2y^{2} + 7y + 6 = 0$

The roots of $2 k x^{2} + 5 k x + 2 = 0$ are equal if $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module