Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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Solving the original equation, we get <math>x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.</math> ~[[User:emerald_block|emerald_block]] | Solving the original equation, we get <math>x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.</math> ~[[User:emerald_block|emerald_block]] | ||
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+ | ==Solution 4== | ||
+ | This is basically bashing using Vieta's formulas (which I highly do not recommend, I only wrote this solution for fun). | ||
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+ | We use Vieta's to find a quadratic relating <math>x</math> and <math>y</math>. We set <math>x</math> and <math>y</math> to be the roots of the quadratic <math>Q ( n ) = n^2 - 4n - 2</math> (because <math>x + y = 4</math>, and <math>xy = -2</math>). We can solve the quadratic to get the roots <math>2 + \sqrt{6}</math> and <math>2 - \sqrt{6}</math>. <math>x</math> and <math>y</math> are "interchangeable", meaning that it doesn't matter which solution <math>x</math> or <math>y</math> is, because it'll return the same result when plugged in. So we plug in <math>2 + \sqrt{6}</math> for <math>x</math> and <math>2 - \sqrt{6}</math> and get <math>\boxed{\textbf{D}\ 440}</math> as our answer. | ||
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+ | ~Baolan | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:38, 1 February 2020
Real numbers and satisfy and . What is the value of
Solution 1
Continuing to combine From the givens, it can be concluded that . Also, This means that . Substituting this information into , we have . ~PCChess
Solution 2
As above, we need to calculate . Note that are the roots of and so and . Thus where and as in the previous solution. Thus the answer is .
Solution 3
Note that Now, we only need to find the values of and
Recall that and that We are able to solve the second equation, and doing so gets us Plugging this into the first equation, we get
In order to find the value of we find a common denominator so that we can add them together. This gets us Recalling that and solving this equation, we get Plugging this into the first equation, we get
Solving the original equation, we get ~emerald_block
Solution 4
This is basically bashing using Vieta's formulas (which I highly do not recommend, I only wrote this solution for fun).
We use Vieta's to find a quadratic relating and . We set and to be the roots of the quadratic (because , and ). We can solve the quadratic to get the roots and . and are "interchangeable", meaning that it doesn't matter which solution or is, because it'll return the same result when plugged in. So we plug in for and and get as our answer.
~Baolan
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.