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Daily Magic Spell Discussion - January 5, 2017

 
 
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Daily Magic Spell Discussion - January 5, 2017
by Areteem Institute - Tuesday, 14 February 2017, 10:19 AM
 

For today, we will discuss the problem given on January 5, 2017:

Problem: Let $x_1, x_2$ be the two roots for equation $x^2+x-3=0$, find the value of $x_1^3-4x_2^2+19$.

Solution: $x_1,x_2$ are roots of the equation $x^2+x-3$, so $x_1^2+x_1-3=0$ and similarly $x_2^2+x_2-3=0$.  Hence, $x_1^2=3-x_1,x_2^2=3-x_2$, and we have $x_1^3-4x_2^2+19=x_1(3-x_1)-4(3-x_2)+19$. Simplifying and substituting again we have $3x_1-x_1^2+4x_2+7=3x_1-(3-x_1)+4x_2+7=4(x_1+x_2)+4$. By Viete's Theorem we have $x_1+x_2=-1$, so $4(x_1+x_2)+4 = 4(-1)+4 = 0$.

For some students, a possible approach to this problem is to naively determine the roots of $x^2 + x - 3 = 0$ and plug the values of the roots into $x_1^3-4x_2^2+19$, making the process extremely vulnerable to algebraic mistakes. Because of this, there are various answers that can be given depending on the algebraic mistake that was produced by the student.

There is a possibility that the student incorrectly remembers Viete's Formula. They may have forgotten that there is a negative sign for the sum of the two roots and because of this $x_1 +  x_2 = -1$.

If you have any approach that you would like to share, please post below!