Question

If one root of the quadratic equation kx² – 5x + 2 = 0 is 4 times the other, find k.

Answer 1

Given: kx² – 5x + 2 = 0
On comparing this equation with ax² + bx + c = 0, we get:
a = k, b = –5 and c = 2
Let α and β be the roots of the quadratic equation kx² – 5x + 2 = 0. Then,
α = 4 β
We know that α + β = –$\frac{b}{a}$ and αβ = $\frac{c}{a}$.
From α + β = –$\frac{b}{a}$, we get:
4 β + β = $-\left(\frac{-5}{k}\right)=\frac{5}{k}$
$\Rightarrow$5β = $\frac{5}{k}$
$\Rightarrow$β = $\frac{1}{k}$ …(1)
From αβ = $\frac{c}{a}$, we get:
4 β $\times$β = $\frac{2}{k}$
$\Rightarrow$4 β² = $\frac{2}{k}$

$\Rightarrow$ β² = $\frac{1}{2k}$
From the equation (1), we get:

$$\begin{gathered} {\left(\frac{1}{k}\right)}^2=\frac{1}{2k} \\ \Rightarrow \frac{1}{k^2}=\frac{1}{2k} \\ \Rightarrow k^2=2k \\ \Rightarrow k^2-2k=0 \\ \Rightarrow k(k-2)=0 \\ \Rightarrow k=0or2 \end{gathered}$$

Now, on substituting k = 0 in the given equation, we observe that the coefficient of x² becomes zero, which is not possible for a quadratic equation. Thus, k = 0 is not possible.
Therefore, k = 2.