If l, m, n are real and l neq m,then the root | vedclass

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

Question

(B) The given quadratic equation is $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$.The discriminant $D$ of a quadratic equation $ax^2 + bx + c = 0$ is given

Vedclass · Accepted Answer

Real and distinct

Vedclass · Answer

(B) The given quadratic equation is $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$.The discriminant $D$ of a quadratic equation $ax^2 + bx + c = 0$ is given by $D = b^2 - 4ac$.Here,$a = (l - m)$,$b = -5(l + m)$,and $c = -2(l - m)$.$D = [-5(l + m)]^2 - 4(l - m)(-2(l - m))$$D = 25(l + m)^2 + 8(l - m)^2$.Since $l$ and $m$ are real and $l 
eq m$,$(l - m)^2 > 0$ and $(l + m)^2 \geq 0$.Thus,$D = 25(l + m)^2 + 8(l - m)^2 > 0$.Since the discriminant $D > 0$,the roots of the equation are real and distinct.

If $l, m, n$ are real and $l \neq m$,then the roots of the equation $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$ are

Similar Questions

The point $(t^2 + 2t + 5, 2t^2 + t - 2)$ lies on the line $x + y = 2$ for

Find a polynomial $f(x)$ of degree $2$ where $f(0)=8, f(1)=12, f(2)=18$.

Let $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4 + x^2 + 1 = 0$. Then the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$ is:

If $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,then

Let $S=\{1, 2, 3, \ldots, 100\}$. Suppose $b$ and $c$ are chosen at random from the set $S$. The probability that $4x^2+bx+c=0$ has equal roots is

Vedclass Test Series

Exam Paper Generator

Online Exam Module