2025 AMC 10A Problems/Problem 9

Let . For how many real numbers does the graph of pass through the point ?

(A) (B) (C) (D) (E) more than

Video Solution

https://youtu.be/l1RY_C20Q2M

Solution 1

Substitute for and set this expression equal to The problem boils down to finding how many real roots has. We further simplify this expression and create a function

Using Descarte's Rule of Signs we get:

Sign changes for (possible number of positive roots): 2

Sign changes for (possible number of negative roots): 1

Possibilities for roots:

1) positive roots, negative root

2) positive roots, negative root, imaginary roots

So which one is it? We know if the function changes sign between an interval, then a root exists in that interval. From to the function changes sign because while , so a positive root exists. This eliminates the second possibility, implying that there must be positive and negative roots. So the answer is

~grogg007

Alternative Solution

Since we are trying to find the number of roots of (which is the same as the number of roots of because we divided by a constant, namely ), we can sketch a graph of this function, which we will call . We find that the function is strictly decreasing when , , and the function is strictly increasing when . We immediately see that since is negative, there is a root between and by the Intermediate Value Theorem. From here, we use the first derivative test as follows. Taking the derivative, we have . We solve for the roots of this equation, obtaining . We know that at these two values, the function will either reach a local minimum or maximum if the second derivative isn't . Proceeding by the second derivative rule, we have that is positive at and negative at , making them local minimums and local maximums, respectively. We don't need to care about the latter case because the local maximum doesn't give any more roots besides the one we've already found. Considering the first case when , we must find the sign of . Using some approximations, we have that By the Intermediate Value Theorem again, we see that there is one root between and and one root between and . Thus, we have a total of roots. Therefore, the answer is

~scjh999999

~minor LaTeX edits by i_am_not_suk_at_math

Solution 2

This problem is essentially asking how many values of satisfy since the graph is a shift units right from the original graph of Now we just need to determine the amount of real solutions to

Dividing all terms by 25, we get Using the rational root theorem and testing, we find no rational roots. Our next method to determine the amount of roots is graphing. Notice how when When Substituting gives us Substituting gives us also. This makes us think to see if there was any sudden bump or curve in the graph, so we test the midpoint which is Substituting we get Aha! Connecting the dots, we see there are two solutions from Also, must continue increasing after which tells us there is a third root when Now we can conclude that the problem has 3 real roots, meaning that 3 values of a satisfy the problem. The answer is

~ eqb5000/Esteban Q.

Solution 3 (Graphing)

We can factor as . We just need to check that if the local maximum between and is below or above 25. Observe that , so the graph will cross the line 3 times,

~gimmeaworkingusername

Somewhat cheese (not recommended)

Instead of plugging in a value between and we can assume that the equation will reach a value above between those numbers. Thus, the graph crosses 3 times,

~dodobird150

Super Lazy Solution

Instead of using numbers, we could just remember the shape of the graph given, and that it is extremely vertically dilated. The shape of the graph is(for the y-values given) basically lines, so moving these lines left and right, we can see that the graph will intersect at 3 points, so

~HappyLion

Chinese Video Solution

https://www.bilibili.com/video/BV1bj2uBxEXo/

~metrixgo

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution 2 graphig approach by SpreadTheMathLove

https://www.youtube.com/watch?v=tDclC-ojfuY

Video Solution (In 1 Min)

https://youtu.be/zvB7LuEBbww?si=OJ9H-Wkkx3WUxTU5 ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU?si=MPPNaoCB_Bqo918z&t=1132 ~MK

Video Solution by Daily Dose of Math

https://youtu.be/gPh9w3X3QSw?si=FhsZ3VSzFXKEO5pf&t=749 ~Thesmartgreekmathdude

See Also

• AMC 10
• AMC 10 Problems and Solutions
• Mathematics competitions
• Mathematics competition resources

These problems are copyrighted © by the Mathematical Association of America.