Use the Quadratic Formula to solve an equation of the form ax^2+bx+c=0.
(4,1) and (1,4)
Practice makes perfect
We want to solve the given system of equations using the Substitution Method.
y=x^2-6x+9 & (I) y+x=5 & (II)
The y-variable is isolated in Equation (I). This allows us to substitute its value x^2-6x+9 for y in Equation (II).
Notice that in Equation (II), we have a quadratic equation in terms of only the x-variable.
x^2-5x+4=0 ⇔ 1x^2+( - 5)x+ 4=0We can substitute a= 1, b= - 5, and c= 4 into the Quadratic Formula.
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.
x=5± 3/2
x_1=5+ 3/2
x_2=5- 3/2
x_1=8/2
x_2=2/2
x_1=4
x_2=1
We obtained two values for the x-variable. Now, consider Equation (I).
y=x^2-6x+9
We can substitute x=4 and x=1 into the above equation to find the values for y. Let's start with x=4.
We found that y=1 when x=4. One solution to the system, which is a point of intersection of the parabola and the line, is (4,1). To find the other solution, we will substitute 1 for x in Equation (I) again.
We found that y=4 when x=1. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (1,4).
Checking Our Answer
Checking the answer
We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (4,1). We will substitute 4 and 1 for x and y, respectively, in Equation (I) and Equation (II).
