SOLUTION: Y=(x-5)^2 Y=k In the system of equations above ,k is a constant .IF the system has two points of intersection and the distance between these two points is 10,,what is the value

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Question 1162056: Y=(x-5)^2
Y=k
In the system of equations above ,k is a constant .IF the system has two points of intersection and the distance between these two points is 10,,what is the value of k?
The options are 20, 25 , 30 , 35.
The answer is 25 . I don’t know how to solve it . Kindly help .
Thanks in advance .
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
k=(x-5)^2
+/- sqrt(k)=x-5
x=5+ sqrt (k) and 5- sqrt(k) Those are the points, and the distance is 5+sqrt(k)-(5-sqrt(k))=10
2 sqrt(k)=10, sqrt(k)=5, k=25

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

The quadratic function  Y =  is the parabola.


It has the axis of symmetry x= 5 (vertical line).


The line Y = k is a horizontal line.


The fact that the distance between the intersection  points is 10, means that these points are remoted 5 units 

in both directions from the symmetry axis x= 5.



It, in turns, means that the intersection points are at  x= 0  and  x= 10.


By substituting x= 0 into the quadratic funcrion, you get  Y =  =  = 25.


So, the answer is  k = Y = 25.


You will get the same answer, if you substitute  x= 10  instead of  x= 0 into the quadratic function.


Solved and explained.



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