For $3x^2 + 2kx +k-1 =0$; find the value of $k$ for which the roots of the equation are closest together

Question

For $3x^2 + 2kx +k-1 =0$; find the value of $k$ for which the roots of the equation are closest together

So my first approach to the problem was to find when the discriminant = 0 and then round off or something, however this problem was in the non-calculator section of a worksheet... anyone have any ideas on how to do it?

Answer 1

In $x^2-px+q$, $p$ is the sum and $q$ the product of the roots, and the discriminant $D=p^2-4q$ is the square of the difference of the roots. It seems you want to minimize $|D|$.

Answer 2

I will interpret your question as for an equation $3x^2+2kx+k-1=0$ with two real roots, find $k$ corresponding to the least $|x_1-x_2|$.

Thus, we first calculate the discriminant; this gives that $4k^2-4\times3\times(k-1)\geq0$, but this holds for all $k$.

Thus, we can use the Vieta's theorems.

We observe that $|x_1-x_2|_{min}$ corresponds to $(x_1-x_2)^2_{min}$; but $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=(\frac{2k}{3})^2-4\frac{k-1}{3}$, and the minimum is reached when $k=\frac{3}{2}$.

Hope I helped.

Answer 3

the reduced discriminant of your equation is

$\delta=k^2-3(k-1)$

$=(k-\frac{3}{2})^2+\frac{3}{4}>0$.

the roots $a$ and $b$ are such $|a-b|=\frac{2}{3}\sqrt{\delta}$.

they are closest together if

$\delta$ is minimal, which gives

$k=\frac{3}{2}$.

Answer 4

The absolute difference between the roots of a quadratic equation is $$\frac{\sqrt\Delta}{2|a|}$$

and in your case it suffices to minimize the discriminant,

$$k^2-3k+3.$$ This occurs when $2k-3=0$.