Review question

# If the roots of $2x^2+kx+3=0$, are $\alpha$ and $\beta$, what is $k$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8732

## Solution

The roots of the quadratic equation $2x^2+kx+3=0$, where $k>0$, are $\alpha$ and $\beta$, whilst those of the equation $3x^2-2x+3$ are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$. Calculate the value of $k$.

Since we know the roots of $2x^2+kx+3=0$, we can write this as $2(x-\alpha)(x-\beta)=0.$

After expanding we can compare the coefficients with our original quadratic equation:$2x^2+kx+3=2x^2-2(\alpha+\beta)x+2\alpha\beta.$ We then find

\begin{align} \label{eq:1} k&=-2(\alpha+\beta),\\ \label{eq:2}3&=2\alpha\beta. \end{align}

We can now do the same with the other quadratic, to obtain $3x^2-2x+3=3\left(x-\dfrac{\alpha}{\beta}\right)\left(x-\dfrac{\beta}{\alpha}\right)=3x^2-3\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)x+3$ and hence

\begin{align*} -2&=-3\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)\\ &\implies \dfrac{2}{3} = \dfrac{\alpha^2+\beta^2}{\alpha\beta} \\ &\implies \dfrac{2}{3}\alpha\beta = (\alpha+\beta)^2-2\alpha\beta \\ &\implies (\alpha+\beta)^2 = 4\\ &\implies \dfrac{k^2}{4} = 4\\ &\implies k = 4. \end{align*}

We might note that if we substitute $k=4$ into the original equation, and try to solve it, we find that $\alpha$ and $\beta$ are not real, but complex.