k x + y = 3 kx+y=3 k x + y = 3 y = − x 2 + k y=-x^2+k y = − x 2 + k
In the system of equations above, k k k x y xy x y k k k
See System of Equations for a review of systems.
We first substitute the value of y y y
k x + y = 3
kx+\colorbox{aqua}{$y$}=3 k x + y = 3 y = − x 2 + k
y=-x^2+k
y = − x 2 + k k x − x 2 + k = 3 kx\colorbox{aqua}{$-x^2+k$}=3 k x − x 2 + k = 3 Written in standard form:
− x 2 + k x + k − 3 = 0 -x^2+kx+k-3=0 − x 2 + k x + k − 3 = 0
We can inspect the discriminant of the quadratic (the portion under the square root of the quadratic formula):
Quadratic Formula = − b ± b 2 − 4 a c 2 a \text{Quadratic Formula} = \frac{-b\pm\sqrt{\colorbox{aqua}{$b^2-4ac$}}}{2a} Quadratic Formula = 2 a − b ± b 2 − 4 a c Discriminant = b 2 − 4 a c \text{Discriminant} = b^2-4ac Discriminant = b 2 − 4 a c The cases for the discriminant are as follows:
D > 0 D\gt0 D > 0 D = 0 D=0 D = 0 D < 0 D\lt0 D < 0
− x 2 + k x + k − 3 = 0 \colorbox{aqua}{$-$}x^2+\colorbox{yellow}{$k$}x+\colorbox{chartreuse}{$k-3$}=0 − x 2 + k x + k − 3 = 0 Discriminant = k 2 − 4 ( − 1 ) ( k − 3 ) \text{Discriminant}= k^2-4(-1)(k-3) Discriminant = k 2 − 4 ( − 1 ) ( k − 3 ) Since we want the discriminat to be zero for 1 solution:
0 = k 2 + 4 k − 12 = 0 0 = k^2+4k-12=0 0 = k 2 + 4 k − 12 = 0 0 = ( k + 6 ) ( k − 3 ) 0= (k+6)(k-3) 0 = ( k + 6 ) ( k − 3 ) k = − 6 and 3 k=-6 \text{ and } 3 k = − 6 and 3