Question 1199652: My apologies for posting a question with a typo. Here is what the actual question was:
The number of pairs of integers (x,y) that satisfy both x^2+y^2 ≤ 36 and y = x - 4 is...?
A) 4
B) 5
C) 6
D) 7
E) 8
Thanks so much!
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
x^2+y^2 ≤ 36
x^2+(x-4)^2 ≤ 36 ............... plug in y = x-4
x^2+x^2-8x+16 ≤ 36
2x^2-8x+16 ≤ 36
2x^2-8x+16-36 ≤ 0
2x^2-8x-20 ≤ 0
2(x^2-4x-10) ≤ 0
x^2-4x-10 ≤ 0
Use the quadratic formula to determine that the two roots of x^2-4x-10 = 0 are roughly:
x = -1.7417
x = 5.7417
You can confirm these roots are correct through the use of a graphing calculator (eg: Desmos or GeoGebra)
This means
x^2-4x-10 ≤ 0
leads to
-1.7417 ≤ x ≤ 5.7417
This portion of the parabola is below the x axis.
Because x is an integer, we then say
-1 ≤ x ≤ 5
This provides the following possible values for x:
{-1, 0, 1, 2, 3, 4, 5}
There are 7 items in this set.
They correspond to the 7 different (x,y) integer solutions that satisfy the original inequality and equation.
For instance, x = 3 leads to y = x-4 = 3-4 = -1.
The point (3,-1) is one of the seven integer solutions.
Check:
x^2+y^2 ≤ 36
3^2+(-1)^2 ≤ 36
9+1 ≤ 36
10 ≤ 36
We arrive at a true statement at the end.
That solution (3,-1) has been confirmed with the inequality.
I'll let you check the other 6 solutions.
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Answer: D) 7
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