Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"

Revision as of 09:06, 12 July 2021

Problem

The equation $ax^2 + 5x = 4,$ where $a$ is some constant, has $x = 1$ as a solution. What is the other solution?

Solution

Since $x=1$ must be a solution, $a+5=4$ must be true. Therefore, $a = -1$ . We plug this back in to the original quadratic to get $5x-x^2=4$ . We can solve this quadratic to get $1,4$ . We are asked to find the 2nd solution so our answer is $\boxed{4}$

~Grisham

Solution 2

Plug $x=1$ to get $a=-1$ , so $x^2-5x+4=0$ , or $(x-4)(x-1)=0$ , meaning the other solution is $x=\boxed{4}$ $\linebreak$ ~Geometry285

Solution 3

$ax^2+5x-4=0$ Plugging in $1$ , we get $a+5-4=0 \implies a+1=0 \implies a=-1$ , therefore, $-x^2+5x-4=0 \implies (x-4)(x-1)=0$ Finally, we get the other root is $4$ .

- kante314 -

@@ Line 12: / Line 12: @@
 ~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 -
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