Question

Find k when 2x^2-3kx+5k=0 has one root twice the other

Answer 1

Answer:

k=5

Step-by-step explanation:

take sum of roots and product of roots

equate the 2 equations

Answer 2

Step-by-step explanation:

Let one root be m , then the other root is 2 m.

$\red{ \pmb{\begin{aligned} \therefore \quad \text { The sum of roots } & \tt=m+2 m=\frac{-b}{a} \\ \\ \tt 3 m & \tt=\frac{-(-3 q)}{2}= \frac{3 q}{2} \\ \\ \tt m & \tt=\frac{q}{2} \end{aligned} }}$

and the product of roots $\tt =m \times 2 m=\dfrac{c}{a}$

$\pink{ \pmb{\[ \begin{aligned} \tt 2 m^{2} & \tt=\frac{5 q}{2} \\ \\ \tt 2\left(\frac{q}{2}\right)^{2} &=\frac{5 q}{2} \\ \\ \tt 4 \times \frac{q^{2}}{4} & \tt=5 q \\ \\ \tt q^{2} &=5 q \\ \\ \tt q & \tt=5 \end{aligned} \]}}$