Quadratic equation problem. Composition of functions

Question

Suppose $p(x)$ and $q(x)$ are quadratic polynomials and the three largest roots of $p(q(x))$ are $10$, $20$ and $23$. What is the smallest root of $p(q(x))$?

Then, there will be 4 roots. $q(10)$ $q(20)$ $q(23)$ $q$(another zero) will make the functions equal to zero. The rest...

Calvin Lin · Accepted Answer · 2014-05-24T17:42:02.117

2

Hint: Let the roots of $ p(x) = 0 $ be $\alpha, \beta$.

What can you say are the values of $ q(10), q(20), q(23) $?

They are either $ \alpha$ or $\beta$. Note that at most two of them have the same value, since $q(x)$ is a quadratic function.

Hence, can you conclude what the possibilities of the last root is? (Note, there is more than 1 possibility)

We split into casework, depending on what values they are.
Case 1: $q(10) = \alpha, q (20) = \beta , q (23) = \alpha $. The last solution is $r$ and satisfies $ q(r) = \beta$.
Observe that for any constant $C$, the solutions to $q(x) = C$ sum up to the same value by Vieta's formula. Hence, we have $ 10 + 23 = 20 + r$, which gives $r = 13$.

Complete the rest of the cases.

edited May 24 '14 at 17:42

answered May 22 '14 at 23:57

Calvin Lin

68,864

I don't understand. The values for those function are the zeroes for the composition. – most venerable sir May 23 '14 at 01:40
@user11355 We know that $ p ( q (10) ) = 0 $ right? What does that tell us about $ q(10)$? – Calvin Lin May 23 '14 at 01:41
The y value is the root of p(x) – most venerable sir May 23 '14 at 01:47
They are unversed functions? The y value turns into x value. – most venerable sir May 23 '14 at 01:51
$q(10)= \alpha $ or $\beta$. Do you see why? – Calvin Lin May 23 '14 at 02:14
Yes, i see. Does that help – most venerable sir May 23 '14 at 11:38
Then it's casework. If $q(10) = \alpha, q(20) = \beta, q(23) = \alpha$, what is the last root going to be? – Calvin Lin May 23 '14 at 19:22
Wait, how do you determine which two roots yield the same value . – most venerable sir May 24 '14 at 00:31
As I said, casework. You have 3 cases to deal with. – Calvin Lin May 24 '14 at 03:20
In that case, q(n) is gonna be equal to $\beta$ – most venerable sir May 24 '14 at 13:03
Can we figure out the two functions, if they are equal to the same number? – most venerable sir May 24 '14 at 13:05
No, the functions have an additional degree of freedom to them – Calvin Lin May 24 '14 at 15:05
Then how do we know what $n$ is? – most venerable sir May 24 '14 at 15:06
Do you want be to plug in the values into the quadratic equation? – most venerable sir May 24 '14 at 15:10
No. In the case that I have, the last root of $p(q(x))$ is 13. Do you see why? – Calvin Lin May 24 '14 at 15:12
Yes, I plugged in the values, solved the equation. Since 429a+13b=0... – most venerable sir May 24 '14 at 15:14
Yes. Though, you can simplify it by considering the quadratic graph. 13=23+10-20. Alternatively, use the fact that the sum of roots to $p(x) =C$ is a constant, so 10+23=20+13 – Calvin Lin May 24 '14 at 15:18
Shouldn't it be q(n) +q(10)=q(23)+q(20). If the sum of the two roots is constant. – most venerable sir May 24 '14 at 16:19
What your doing is 2x=2B. But not x+B=x+B. Why it still works? – most venerable sir May 24 '14 at 16:25
@user11355 The sum of roots will be constant by Vieta's formula. Alternatively, they are equal to twice the x-coordiante of the vertex of the (symmetric) parabola. – Calvin Lin May 24 '14 at 17:42

Quadratic equation problem. Composition of functions

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