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MC II-A 3.1 Continued (Quadratics)

 
 
HuAshley的头像
MC II-A 3.1 Continued (Quadratics)
HuAshley - 2023年04月6日 Thursday 22:29
 
In this problem, I wanted to clarify if this solution is a mistake or not. The boxed equation states that root 1 times root 2 is positive q, however, this solution assumes that root 1 times root 2 is negative q, so they get 1 negative and 1 positive root. I got both negative.



 
LensmireJohn的头像
Re: MC II-A 3.1 Continued (Quadratics)
LensmireJohn - 2023年04月10日 Monday 13:57
 

The answer that the quadratic should have exactly one positive root should be correct.

I think you agree that the quadratic always has two real roots, so let's focus on Vieta's theorem for $x^2+px+q=0$. By Vieta's theorem:

  • The sum of the roots is $-p$.
  • The product of the roots is $q$.

Thus, if we know that $p$ is positive, then the sum of the roots is negative, and if $q$ is negative, then the product of the roots is negative. This is only possible if one root is positive and one root is negative.

If it helps, you can visualize the graph using Desmos and check different values of $p$ and $q$ here: www.desmos.com/calculator/e0gxboxui6

Hope this helps! Let us know if you have any further questions.