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Respected All.

I am unable to find out what's so wrong in the following. Please help me.

It is given that $t$ is a common root of the following two equations given by \begin{align} &x^2-bx+d=0 \tag{1}\\ &ax^2-cx+e=0 \tag{2} \end{align} where $a,b,c,d,e$ are real numbers.

Then using cross multiplication technique, we shall get $$\frac{t^2}{cd-be}=\frac{t}{ad-e}=\frac{1}{ab-c} \tag{3}$$ which will give us $$t=\begin{cases} \frac{cd-be}{ad-e},\\ \frac{ad-e}{ab-c},\\ \pm\sqrt{\frac{cd-be}{ab-c}} \end{cases} \tag{I}$$

My problems starts from here. If $t$ satisfies both (1) and (2) then any linear combination of (1) and (2) should be satisfied by $t$. So that by $\lambda \times (1)+\mu \times (2)$ we shall have $$(\lambda+\mu a)t^2-(\lambda b+\mu c)t+(\lambda d+\mu e)=0 \tag{4}$$ where $\lambda, \mu$ are suitable reals for (4) to have real roots.

hence we shall get $$t=\frac{1}{2(\lambda+\mu a)}[(\lambda b+\mu c)\pm \sqrt{(\lambda b+\mu c)^2-4(\lambda+\mu a)(\lambda d+\mu e)}] \tag{II}$$

All three results in (I) and the results in (II) are supposed to be same. Aren't they?

So what if we consider $(a,b,c,d,e)=(3,9,38,14,119)$. Then from (I) we shall get $(7,7,7)$. But when we shall apply (II) then the results are becoming "dirty": if we choose $\lambda=-16,\mu=20$ the results are coming as $(7.057474264\cdots, 7.06\cdots)$

Why is it happening ? Am I making any theoratical mistake?

miracle173
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KON3
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    Could you explain what do you call "cross multiplication technique"? – user37238 Nov 27 '14 at 08:14
  • One other comment: you start with two polynomial equations. The equation (1) has at most two solutions (since the coefficient of the term $x^2$ is $1$). Since you're looking for a solution that satisfies both equation in particular, it should satisfy the first one. So there is at most two solutions... – user37238 Nov 27 '14 at 08:17
  • Do you need the value of t? What exactly has been asked? – ghosts_in_the_code Nov 27 '14 at 09:37

1 Answers1

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We have two polynomial equations $f_1(x)=f_2(x)=0$ with quadratic polynomials $f_1(x)=x^2-bx+d$ and $f_2(x)=ax^2-cx+e$, where $a,b,c,d,e$ are the coefficients. Then $$ t=\frac{ \pm \sqrt{b^2 - 4d} + b}{2} $$ is a common root, if and only if the coefficients satisfy certain poylnomial conditions. To see this, just substitute this to the second equation. We can make a case distinction. Assume that $d=0$. Then $t=b$ is a common root for $b\neq 0$ if $a=\frac{bc - e}{b^2}$, and $c,e$ arbitrary; and for $b=0$ if $e=0$ and $a,b,c$ arbitrary.

If $d\neq 0$, $t=\frac{ \sqrt{b^2 - 4d} + b}{2}$ is a common root if and only if $$ a=\frac{\sqrt{b^2 - 4d}\cdot be - \sqrt{b^2 - 4d}\cdot cd - b^2e + bcd + 2de}{2d^2}. $$ A very similar formula holds for the case $t=-\frac{ \sqrt{b^2 - 4d} + b}{2}$.

For your example $(a,b,c,d,e)=(3,9,38,14,119)$ the formula gives $t=7$. We have $\sqrt{b^2-4d}=5$, so that $t=\frac{ \sqrt{b^2 - 4d} + b}{2}=7$, and the relation between $a,b,c,d,e$ is satisfied.

Edit: For your example $(a,b,c,d,e)=(3,9,38,14,119)$ your equation II gives the solutions $t=7$ and $t=(2\lambda + 17\mu)/(\lambda + 3\mu)$. For $\lambda=-16$ and $\mu=20$ I obtain exactly $t=7$, so no problem. I suppose you have done a computational mistake there.

Dietrich Burde
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  • Dear Sir, as you mentioned about the conditions, what about this: From (I), $\frac{cd-be}{ad-e}=\frac{ad-e}{ab-c}$ ? – KON3 Nov 28 '14 at 08:24
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    Yes, indeed. Everything seems to be correct, but II resp. its solutions for your example are miscalculated - see my edit. – Dietrich Burde Nov 28 '14 at 12:10
  • Sir, you are right. I did a mistake in my calculation. Thank you so much for this. – KON3 Dec 01 '14 at 08:15