If ax^2 + bx + c = 0,then x = | vedclass

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

Question

(C) The quadratic formula for the equation $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Let us check option $(C)$: $\frac{2c}{-b \m

Vedclass · Accepted Answer

$\frac{2c}{-b \mp \sqrt{b^2 - 4ac}}$

Vedclass · Answer

(C) The quadratic formula for the equation $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Let us check option $(C)$: $\frac{2c}{-b \mp \sqrt{b^2 - 4ac}}$.Rationalizing the denominator:$\frac{2c}{-b - \sqrt{b^2 - 4ac}} 	imes \frac{-b + \sqrt{b^2 - 4ac}}{-b + \sqrt{b^2 - 4ac}} = \frac{2c(-b + \sqrt{b^2 - 4ac})}{b^2 - (b^2 - 4ac)} = \frac{2c(-b + \sqrt{b^2 - 4ac})}{4ac} = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$.Similarly,$\frac{2c}{-b + \sqrt{b^2 - 4ac}} = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$.Thus,the expression in option $(C)$ is equivalent to the standard quadratic formula.

If $ax^2 + bx + c = 0$,then $x =$

Similar Questions

If $a, b, c$ are in Arithmetic Progression $(AP)$,then the roots of the equation $ax^2 - 2bx + c = 0$ are

Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then,for any choice of positive integers $a$ and $b$,$t^3$ is never equal to:

The $x$-coordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3|x| + 2 = 0$ and the $y$-coordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$. Then the possible vertices of the square are:

Assertion $(A)$: The maximum value of $-x^2+3x+1$ is $\frac{13}{4}$.
Reason $(R)$: If $a < 0$,the maximum value of $ax^2+bx+c$ exists at $x = -\frac{b}{2a}$.
The correct option among the following is

The sum of all the roots of the equation $(x-1)^{2}-5|x-1|+6=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module