SOLUTION: The parabola y=x2-4 intersects the x axis at two points, P and Q. What is the distance from P to Q?

Algebra ->  Expressions-with-variables -> SOLUTION: The parabola y=x2-4 intersects the x axis at two points, P and Q. What is the distance from P to Q?      Log On


   

Question 120433: The parabola y=x2-4 intersects the x axis at two points, P and Q. What is the distance from P to Q?
Found 2 solutions by josmiceli, bucky:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+x%5E2+-+4
y+=+%28x+%2B+2%29%28x+-+2%29
set y+=+0 to find roots
0+=+%28x+%2B+2%29%28x+-+2%29
The x-intercepts are the values of x that solve this
x+=+2
x+=+-2
So, P+=+-2 and Q+=+%2B2
The distance apart is Q+-+P+=+2+-%28-2%29
Q+-+P+=+4 answer

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given: the parabola
.
y+=+x%5E2-4
.
To find the intersections with the x-axis, set y = 0 because any point on the x-axis has
zero as its y-value. So the equation to solve then becomes:
.
0+=+x%5E2-4
.
Solve this by getting rid of the -4 on the right side by adding +4 to both sides. On the right
side the -4 and the +4 cancel each other when they are added. So when you add +4 to both sides
the equation becomes:
.
4+=+x%5E2
.
Solve for x by taking the square root of both sides to get two answers:
.
x+=+sqrt%284%29=2 and x+=+sqrt%284%29+=+-2
.
So the intersections of the parabola and the x-axis occur on the x-axis at x = -2 and at
x = +2. These points are P and Q. And, if you subtract them you get:
.
2-%28-2%29+=+2%2B2+=+4
.
The points P and Q are 4 units apart.
.
You can see this in the graph of the parabola shown below:
.
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2Cx%5E2+-+4%29
.
Hope this helps you to understand the problem.