If both the roots of 25x^{2} - x(m - 2) - 1 = | vedclass

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

Question

(C) Let the roots of the quadratic equation $25x^{2} - x(m - 2) - 1 = 0$ be $\alpha$ and $-\alpha$ because they are opposite to each other.For a quadr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let the roots of the quadratic equation $25x^{2} - x(m - 2) - 1 = 0$ be $\alpha$ and $-\alpha$ because they are opposite to each other.For a quadratic equation $ax^{2} + bx + c = 0$,the sum of roots is given by $\alpha + \beta = -b/a$.Here,$a = 25$,$b = -(m - 2)$,and $c = -1$.Sum of roots: $\alpha + (-\alpha) = -[-(m - 2)] / 25$.$0 = (m - 2) / 25$.Multiplying both sides by $25$,we get $0 = m - 2$.Therefore,$m = 2$.

If both the roots of $25x^{2} - x(m - 2) - 1 = 0$ are opposite,then $m = \ldots$

Similar Questions

Find the roots of the following quadratic equation using the quadratic formula,if they exist: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$; $(x \neq -1, -2, -4)$

If the roots of the following quadratic equation exist,find them by the method of completing the square: $\frac{2}{x^{2}}-\frac{1}{x}=6$.

Examine whether the following equation is quadratic or not: $x^{3}-4x^{2}-x+1=(x-2)^{3}$

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $2x^{2} - 6x + 3 = 0$; $x = \frac{1}{2}$.

The value of discriminant $D$ for the equation $(3x - 14)^2 = 0$ is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module