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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.

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    Shame on me, I need to be more patient. Sorry2012-01-15

5 Answers 5

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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.

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    @user1131662: You certainly were right. After a serious attempt, help is part of the process of learning.2012-01-16
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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.

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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 $a=1/2$.

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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.

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    Put $a=1/2$ into $\frac{a+3-\sqrt{a^2-18a+9}}{8}$, we get $\frac{a+3-\sqrt{a^2-18a+9}}{8}=\frac{3}{8}$, which is not a real solution.2012-01-15
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The resultant of the two polynomials is $\left| \begin{array} {cccc} 2 & -3 & 1 & 0 \\ 0 & 2 & -3 & 1 \\ 2 & -a-3 & 3a & 0 \\ 0 & 2 & -a-3 & 3a \end{array} \right| .$

Equating this determinant to $0$ we get $2 (2a-1) (5a-2)=0$, giving $a = \frac 1 2$ and $a = \frac 2 5$.

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    Partial duplicate of Robert Israel's answer as well.2015-08-17