find the least a, for which two equations have a common root

Question

Could you help me out please. I have two equations: $2x^2-3x+1=0 $ and $ 2x^2-(a+3)x+3a=0$

I need to find the least $a$ for which these two equations have a common root. At a first glance I thought it'd be easy, just creating an equation with these two, then creating a function for $a$ and then just a small derivative knowledge. But, unfortunately it seems that it's not as simple as I think, because I've been getting very strange answers.

It'd be wonderful if you could help me out here, just can't concentrate enough maybe.

Answer 1

No derivative knowledge is needed. A common zero of your two polynomials is a root of their difference. So it must be a root of the equation $ax=3a-1$. It is easy to see that $a\ne 0$. So any common root must be equal to $3-1/a$. Substitute in the first equation and solve for $a$.

Because this is homework, we omit the rest of the calculation. But after a while you should get a quadratic in $a$.

Comment: For various reasons, it is nice to put off dividing as long as possible. Since $a\ne 0$, we can rewrite the first equation as $a^2x^2-3a^2x+a^2=0$. Then we can substitute $3a-1$ for $ax$. This yields $$2(3a-1)^2 -3a(3a-1)+a^2=0,$$ and then simplification is pleasant and quick.

Answer 2

I think you can just do it in a concrete way. The roots of $2x^2-3x+1=0$ are $\frac{1}{2}\mbox{ and } 1.$ By quadratic formula, the roots of $2x^2-(a+3)x+3a=0$ are $$\frac{a+3-\sqrt{a^2-18a+9}}{8}\mbox{ and }\frac{a+3+\sqrt{a^2-18a+9}}{8}.$$ If they have common roots, we have the following possibilities: $$\frac{a+3-\sqrt{a^2-18a+9}}{8}=\frac{1}{2}\mbox{ or }\frac{a+3-\sqrt{a^2-18a+9}}{8}=1,$$ or $$\frac{a+3+\sqrt{a^2-18a+9}}{8}=\frac{1}{2}\mbox{ or }\frac{a+3+\sqrt{a^2-18a+9}}{8}=1.$$ Each one of them is a quadratic equation in $a$ which can be solved.

Answer 3

Note that these equation cannot have two common roots. Because if they did, then one equation is a scalar multiple of the other, in which case, $$\dfrac{2}{2}=\dfrac{-3}{-(a+3)}=\dfrac{1}{3a}$$ The first equality would mean that, $3=a+3 \implies a=0$ which makes the second equality absurd.

So, let $\beta$ be the common root of these two equations. We have,

$$2\beta^2-3\beta+1=0$$ $$2\beta^2-(a+3)\beta+3a=0$$ Now we use the Cramer's rule to observe that,

We have that, $$ \dfrac{\beta^2}{\left| \begin{array}{rr} -3 & 1 \\ -(a+3) & 3a \end{array} \right| }=\dfrac{\beta}{\left| \begin{array}{rr} 1 & 3a \\ 2 & 2 \end{array} \right|}=\dfrac{1}{\left| \begin{array}{rr} 2 & 2 \\ -3 & -(a+3) \end{array} \right|}$$

This gives you,

$$\dfrac{\beta^2}{-8a+3}=\dfrac{\beta}{2-6a}=\dfrac{1}{-2a}$$

This yields, on eliminating $\beta$, $$\left(\dfrac{2-6a}{-2a}\right)^2=\dfrac{-8a+3}{-2a}$$ This simplifies to the following, $$10a^2-9a+2=0$$ whose roots are $\dfrac{2}{5}$ and $\dfrac{1}{2}$ which implies, the least $a$ is $\dfrac{2}{5}$. So, this completes your answer.

Answer 4

The general theory is this: two polynomials have a common root when their resultant is $0$. There are various ways to calculate that. The resultant of $2 x^2 - 3 x + 1$ and $2 x^2-(a+3) x+3 a$ is $20 a^2-18 a+4 = 2 (5 a - 2) (2 a - 1)$, so there is a common root for $a=2/5$, and $a=1/2$.