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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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This particular problem can be done through ad hoc methods. E.g., why can't you find the roots of the first equation and plug them in the second? Each of the two roots gives a different value of $a$. –  Srivatsan Jan 15 '12 at 22:59
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I agree with Srivatsan. There's no need for derivatives here. –  Dylan Moreland Jan 15 '12 at 23:01
    
Oh, I'm really sorry to bother you, just didn't think it'd be so simple. –  user1131662 Jan 15 '12 at 23:08
    
Shame on me, I need to be more patient. Sorry –  user1131662 Jan 15 '12 at 23:09

4 Answers 4

up vote 5 down vote accepted

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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Yes, or just inverting this one, calculating the roots in the first equation and substituting instead of $x$ in the second. I wonder, how couldn't I do it by myself –  user1131662 Jan 15 '12 at 23:16
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Maybe thinking "What tools shall I use" instead of "What is the problem?". –  André Nicolas Jan 15 '12 at 23:21
    
Maybe, or maybe not. However I tried to solve this in many ways, some of them are below. But I got stuck a few times so decided to ask for a hint. Wasn't I right? –  user1131662 Jan 16 '12 at 12:47
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@user1131662: You certainly were right. After a serious attempt, help is part of the process of learning. –  André Nicolas Jan 16 '12 at 15:17

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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Yes, this was my first guess, but I was astonished with $a^2-18a+9$ because couldn't find roots without a calculator, so I thought something's not right. –  user1131662 Jan 15 '12 at 23:12
    
Wait! I was wrong. It's linear equation instead of quadratic equation! Look at the first equation for example, we have $\sqrt{a^2-18a+9}=a-1$, which implies that $a^2-18a+9=a^2-2a+1$, or equivalently, $16a=8$, or $a=1/2$. Of course we need to put it back to check this is a real solution since we have taken square. –  Paul Jan 15 '12 at 23:23
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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. –  Paul Jan 15 '12 at 23:25

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