Question

If $l,m$ are real,$l\ne m$, then the roots by the equation:$(l-m)x^2-5(l+m)x-2(l-m)=0$are

• A. $x>y$
• B. $x<y$
• C. $x=y$
• D. real and unequal

Answer 1

The correct option is D

real and unequal

Explanation for the correct answer:

$(l-m)x^2-5(l+m)x-2(l-m)=0$ is the given equation.

Comparing given equation with standard form of quadratic equation $a{x}^2+bx+c=0$

we get $a=l-m$, $b=-5\left(l+m\right)$ , $c=-2\left(l-m\right)$

The nature of the roots of a quadratic equation depends on the value of $D$, where $D$is given as

$D=b^2-4ac$

Substituting the values of $a,b,c$ for the given equation we get

$\Rightarrow$$D=5^2{\left(l+m\right)}^2-4\left(l-m\right)\left(-2\right)\left(l-m\right)$

$\Rightarrow$$D=25{\left(l+m\right)}^2+8{\left(l-m\right)}^2$

For all real values of $l,m$${\left(l+m\right)}^2>0,{\left(l-m\right)}^2>0$ as $l\ne m$

$\Rightarrow$$D>0$

When $D>0$ for a quadratic equation , the equation has real and unequal roots.

Hence, the given quadratic equation $(l-m)x^2-5(l+m)x-2(l-m)=0$ has real and unequal roots,

Hence, option(D) is the correct answer.