Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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== Solution 2 == | == Solution 2 == | ||
As above, we need to calculate <math>\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}</math>. Note that <math>x,y,</math> are the roots of <math>x^2-4x-2</math> and so <math>x^3=4x^2+2x</math> and <math>y^3=4y^2+2y</math>. Thus <math>x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88</math> where <math>x^2+y^2=20</math> and <math>x^2y^2=4</math> as in the previous solution. Thus the answer is <math>\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}</math>. | As above, we need to calculate <math>\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}</math>. Note that <math>x,y,</math> are the roots of <math>x^2-4x-2</math> and so <math>x^3=4x^2+2x</math> and <math>y^3=4y^2+2y</math>. Thus <math>x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88</math> where <math>x^2+y^2=20</math> and <math>x^2y^2=4</math> as in the previous solution. Thus the answer is <math>\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}</math>. | ||
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== Solution 3 == | == Solution 3 == |
Revision as of 11:15, 4 June 2020
Contents
Problem
Real numbers and satisfy and . What is the value of
Solution
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 (Bashing)
This is basically bashing using Vieta's formulas to find and (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
Solution 5 (Bashing Part 2)
This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.
We first change the original expression to , because . This is equal to . We can factor and reduce to . Now our expression is just . We factor to get . So the answer would be .
Solution 6 (Complete Binomial Theorem)
We first simplify the expression to Then, we can solve for and given the system of equations in the problem. Since we can substitute for . Thus, this becomes the equation Multiplying both sides by , we obtain or By the quadratic formula we obtain . We also easily find that given , equals the conjugate of . Thus, plugging our values in for and , our expression equals By the binomial theorem, we observe that every second terms of the expansions and will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of . Thus, our expression equals which equals which equals .
~ fidgetboss_4000
Solution 7
As before, simplify the expression to Since and , we substitute that in to obtain Now, we must solve for . Start by squaring , to obtain Simplifying, . Squaring once more, we obtain Once again simplifying, . Now, to obtain the fifth powers of and , we multiply both sides by . We now have , or We now solve for . , so . Plugging this back into, we find that , so we have . This equals 440, so our answer is .
~Binderclips1
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.