Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | Since <math>x=1</math> must be a solution, <math>a+5=4</math> must be true. Therefore, <math>a = -1</math>. We plug this back | + | Since <math>x=1</math> must be a solution, <math>a+5=4</math> must be true. Therefore, <math>a = -1</math>. We plug this back into the original quadratic to get <math>5x-x^2=4</math>. We can solve this quadratic to get <math>1,4</math>. We are asked to find the 2nd solution so our answer is <math>\boxed{4}</math> |
~Grisham | ~Grisham | ||
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- kante314 - | - kante314 - | ||
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+ | == Solution 4 == | ||
+ | |||
+ | We can rearrange the equation to get that <math>ax^2 + 5x - 4 = 0</math>. Then, by Vieta's Formulas, we have <cmath>x = -\frac{4}{a}</cmath> and <cmath>1+x = -\frac{5}{a},</cmath> where <math>x</math> is the second root of the quadratic. Solving for <math>x</math> tells us that the answer is <math>\boxed{4}</math>. | ||
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+ | ~Mathdreams | ||
==See also== | ==See also== |
Latest revision as of 11:42, 12 July 2021
Problem
The equation where
is some constant, has
as a solution. What is the other solution?
Solution
Since must be a solution,
must be true. Therefore,
. We plug this back into the original quadratic to get
. We can solve this quadratic to get
. We are asked to find the 2nd solution so our answer is
~Grisham
Solution 2
Plug to get
, so
, or
, meaning the other solution is
~Geometry285
Solution 3
Plugging in
, we get
, therefore,
Finally, we get the other root is
.
- kante314 -
Solution 4
We can rearrange the equation to get that . Then, by Vieta's Formulas, we have
and
where
is the second root of the quadratic. Solving for
tells us that the answer is
.
~Mathdreams
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.