2015 AMC 10A Problems/Problem 16
Contents
Problem
If , and , what is the value of ?
Solutions
Solution 1
Note that we can add the two equations to yield the equation
Moving terms gives the equation
We can also subtract the two equations to yield the equation
Moving terms gives the equation
Because we can divide both sides of the equation by to yield the equation
Substituting this into the equation for that we derived earlier gives
Solution 2 (Algebraic)
Subtract from the left hand side of both equations, and use difference of squares to yield the equations
and .
It may save some time to find two solutions, and , at this point. However, in these solutions.
Substitute into .
This gives the equation
which can be simplified to
.
Knowing and are solutions is now helpful, as you divide both sides by . This can also be done using polynomial division to find as a factor. This gives
.
Because the two equations and are symmetric, the and values are the roots of the equation, which are and .
Squaring these and adding them together gives
.
Alternative Ending
Note that . Using Vieta's, we find that and . Subtracting them gives .
~Andrew2019
Solution 3
By graphing the two equations on a piece of graph paper, we can see that the point where they intersect that is not on the line is close to the point (or ). , and the closest answer choice to is .
Note
This is risky, as could be a viable answer too. Do not use this method unless you're sure about the answer. In other words, this solution is less reliable than the others, so only use it if you can't do the other methods.
Solution 4 (When you can't algebraically manipulate)
This, by the way, is a really cheap way of solving the problem but as long as you get an answer, it doesn't matter.
Looking at the first given equation, begin searching for solutions. Notice how works but when plugged into the
second equation, you get . Now, if you decrease the value by to obtain , plugging it into
the second equation will yield . Now, notice how the LHS is now less than the RHS as opposed to what it had
been when we plugged into the second given equation. We then conclude that there must be a solution between
and , so calculating of and we obtain
and
Thus we know the answer to be .
This is a really bad solution, but it works. :/
Solution 5 (Very in depth and reasoned)
Note: In this solution, I explain my entire thought process and the reason why I did each step, so you can read a more intuitive solution rather than a contrived one.
We first notice that each equation has a , which inspires us to isolate the on both equations and equate them to see if we can gather any useful information on and
Doing this we get Now you might be tempted to expand, but notice how we have two perfect squares in our equation: and which motivates us to take advantage of the difference of squares. For simplicity, we suppose that and Thus our equation is now We manipulate this equation by getting the LHS in a form that lets us do difference of squares, so we can rewrite as
Now we can finally use our difference of squares to get
Let's now substitute and back to get information solely for and
Doing this we get,
Now let us divide on both sides and clean up the equation. When we do this, we get that
We got which means that this is fact is only of use if we can obtain it in another way, so we add the LHS and the RHS of both equations and equate them to make our fact of essence.
Doing that, we get Also, when we set this equation up we not only do we make our fact useful, but we also can see that after expanding the RHS, we get on the RHS too, which makes it even a better move that we added both equations because remember our goal is to find
Expanding and plugging in we get so we can plug that in and we get or after cleaning up some algebra.
~triggod
Solution 6 (Quickest)
Subtract the two equations to get . Plugging back into any of the original equations yields . However, we know . Then, we can just multiply the first by two and add to get
Video Solutions
Video Solution by OmegaLearn
https://youtu.be/0W3VmFp55cM?t=15
~ pi_is_3.14
Video Solution 2
~savannahsolver
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.