Question

If one root of the equation $l{x}^2+mx+n=0$ is $\frac{9}{2}$ (where,$l,m,\text{and}n$ are positive integers and $\frac{\mathrm{m}}{4\mathrm{n}}=\frac{\mathrm{l}}{\mathrm{m}}$, then $l+n$ is equal to

• A. $80$
• B. $85$
• C. $90$
• D. $95$
• E. $100$

Answer 1

The correct option is B

$85$

Explanation for the correct solution

Step 1: Information required for the solution

The discriminant $\left(D\right)$ of a quadratic equation tells us about the nature of the roots of the equation.

When $D<0$, the roots will be imaginary, when $D=0$, the roots will be equal, and when $D\ge 0$, then the roots will be real.

The discriminant of an equation $a{x}^2+bx+c=0$ is given by

$\mathbf{\therefore }\mathit{D}\mathbf{=}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathit{a}\mathit{c}$

Step 2: Calculation of the value of $\mathit{l}\mathbf{+}\mathit{n}$.

Here, the given equation is $l{x}^2+mx+n=0$ then its discriminant will be

$\therefore D=m^2-4\ln$

It is also given that $\frac{\mathrm{m}}{4\mathrm{n}}=\frac{\mathrm{l}}{\mathrm{m}}$

After the cross multiplication, we can see that

$\therefore m^2-4\ln =0$

This assures that the discriminant of the equation is $0$. So, the equation has equal roots.

Now, one of the roots of the equation is $\frac{9}{2}$ and the other will be the same.

Then quadratic equation will be ${\left(x-\frac{9}{2}\right)}^2=0$ which after expansion will be

$\begin{array}{rcl}\therefore x^2+\frac{81}{4}-9x& =& 0\\ & \Rightarrow & 4x^2+81-36x=0\end{array}$

Compare the equation $l{x}^2+mx+n=0$ with $\begin{array}{rcl}4x^2+81-36x& =& 0\end{array}$ will give $l=4,m=-36,\text{and}n=81$, then

$\begin{array}{rcl}\therefore l+n& =& 4+81\\ & =& 85\end{array}$

Hence, the correct option is (B).