Quadratic Formula

[Colacanth]

There are two quadratics, both using \(\displaystyle x\) as their variable.

• When both quadratics are evaluated at \(\displaystyle x=1\) their results are the same.
• When both quadratics are evaluated at \(\displaystyle x=2\) their results are the same.
• When both quadratics are evaluated at \(\displaystyle x=3\) their results are the same.

Are the quadratics necessarily the same?

Hint: Suppose the common results from the quadratics are \(\displaystyle k_1\) at \(\displaystyle x=1\), \(\displaystyle k_2\) at \(\displaystyle x=2\), and \(\displaystyle k_3\) at \(\displaystyle x=3\). Can you find the coefficients of the first quadratic in terms of the \(\displaystyle k_i\)'s?

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Following the hint, I arrive at \(\displaystyle (-k_3+3k_2-2k_1)x^2+(3k_3-8k_2+5k_1)x-(2k_1+5k_2-2k_3)\) for the first quadratic. I'm inclined to say that yes, they are the same, but the wording of the question makes me feel otherwise.

Could someone show me a counterexample (if possible) and how it would be reached from the question, please?

 

[Colacanth]

Alright, thanks!

 

[HallsofIvy]

There are two quadratics, both using \(\displaystyle x\) as their variable.

• When both quadratics are evaluated at \(\displaystyle x=1\) their results are the same.
• When both quadratics are evaluated at \(\displaystyle x=2\) their results are the same.
• When both quadratics are evaluated at \(\displaystyle x=3\) their results are the same.

Are the quadratics necessarily the same?

Hint: Suppose the common results from the quadratics are \(\displaystyle k_1\) at \(\displaystyle x=1\), \(\displaystyle k_2\) at \(\displaystyle x=2\), and \(\displaystyle k_3\) at \(\displaystyle x=3\). Can you find the coefficients of the first quadratic in terms of the \(\displaystyle k_i\)'s?

-------------------------

Following the hint, I arrive at \(\displaystyle (-k_3+3k_2-2k_1)x^2+(3k_3-8k_2+5k_1)x-(2k_1+5k_2-2k_3)\) for the first quadratic. I'm inclined to say that yes, they are the same, but the wording of the question makes me feel otherwise.

Could someone show me a counterexample (if possible) and how it would be reached from the question, please?

I'm not sure how you arrived at that! But if you start by writing the two quadratics as \(\displaystyle ax^2+ bx+ c\) and \(\displaystyle \alpha x^2+ \beta x+ \gamma\) then at x= 1 we have \(\displaystyle a+ b+ c= \alpha+ \beta+ \gamma\), at x= 2, \(\displaystyle 4a+ 2b+ c= 4\alpha + 2\beta+ \gamma\), and \(\displaystyle 9a+ 3b+ c= 9\alpha+ 3\beta+ \gamma\).

If we subtract the second equation from the third, both c and \(\displaystyle \gamma\) cancel and we get \(\displaystyle 5a+ b= 5\alpha+ \beta\). If we subtract the first equation from the second, again both c and \(\displaystyle \gamma\) cancel and we get \(\displaystyle 3a+ b= 3\alpha+ \beta\). Now subtract the first of those two equations from the other so that b and \(\displaystyle \beta\) cancel and we have \(\displaystyle 2a= 2\alpha\) so that \(\displaystyle a= \alpha\). Then we can cancel a and \(\displaystyle \alpha\) to have \(\displaystyle b= \beta\), and can cancel both those and a and \(\displaystyle \alpha\) in the first equation to get \(\displaystyle c= \gamma\).