# 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 | ||

+ | |||

+ | ==Solution 2== | ||

+ | Plug <math>x=1</math> to get <math>a=-1</math>, so <math>x^2-5x+4=0</math>, or <math>(x-4)(x-1)=0</math>, meaning the other solution is <math>x=\boxed{4}</math> | ||

+ | <math>\linebreak</math> | ||

+ | ~Geometry285 | ||

+ | |||

+ | == Solution 3 == | ||

+ | <cmath>ax^2+5x-4=0</cmath>Plugging in <math>1</math>, we get <math>a+5-4=0 \implies a+1=0 \implies a=-1</math>, therefore, | ||

+ | <cmath>-x^2+5x-4=0 \implies (x-4)(x-1)=0</cmath>Finally, we get the other root is <math>4</math>. | ||

+ | |||

+ | - kante314 - | ||

+ | |||

+ | == 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>. | ||

+ | |||

+ | ~Mathdreams | ||

==See also== | ==See also== | ||

− | #[[2021 JMPSC | + | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] |

− | #[[2021 JMPSC | + | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] |

#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||

{{JMPSC Notice}} | {{JMPSC Notice}} |

## 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.