Quadratic difficulty
Algebra
Level
5
Given a quadratic equation with integral coefficients y=ax^2 +bx +c and a is co-prime to b and c. It has x and y intercepts at (b,0) and (0,c) respectively and one of its roots is -2.
Find y if x=0 by using the equation that will give the smallest value of y.
The answer is -8.
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1 solution
We could never have a solution to this if a is not 1 or -1.
We know that b is one of the roots. Let r be the other root. Then the equation will be in the form
y= (x-r)(x-b) or
y=(r-x)(x-b)
y=(x-r)(b-x)
one of the roots is -2. There are two options. If b=-2 or r=-2
if b=-2
(1st) y=(x-r)(x+2) = x^2 +(2-r)x -2r. This implies, b=(2-r) => r=4. Then y=(x-4)(x+2)= x^2 -2x -8
(2nd) y=(r-x)(x+2) =-x^2 +(r-2)x +2r. this implies b=r-2 => r=0. Then y=-x^2 -2x
If r=-2
(3rd) y=(x+2)(x-b)= x^2 +(2-b)x -2b. this implies b=2-b => b=1. We have y=x^2+x-2
(4th) y=(x+2)(b-x) = -x^2 +(b-2)x +2b. this implies, b=b-2 . no solution
So we have three equations. At x=0 , we have
y= x^2 -2x -8 --> -8
y=-x^2 -2x --> 0
y= x^2+x-2 --> -2
answer -8