EXAMPLE QUESTION FOR FINDING INTERSECTION

Find intersection
Find intersection of y equals x square minus 1 and y square plus x square equals 30.
Solving:
y square plus x square equals 30 is a circle has a radius of square root of 30 and centered in point (0,0) and y equals x square minus 1 is quadratic equation has vertex of -1. For to find a point of intersection we must to substitution y equal x square minus 1 to equation of y square plus x square equals 30.
y equals x square minus 1 become y plus 1 equals x square then this equation we substitution to y square plus x square equals 30, become:
y square plus y plus 1 equals 30, then 30 we transfer to left articulation become:
y square plus y plus 1 minus 30 equals zero
y square plus y minus 29 equals zero
Then, we find y variable with abc formula, become:
y equals minus 1 plus minus square root of open bracket 1 minus 4 times 1times minus 29 close bracket all over 2
y equals minus 1 plus minus square root of 117 all over 2,
so, y1 equals minus 1 plus square root of 117 all over 2
y1 equals minus 1 plus 10 point 81 all over 2 equals 9 point 81 all over 2 equals 4 point 905
y2 equals minus 1 minus square root of 117 all over 2
y2 equals minus 1 minus 10 point 81 all over 2 equals minus 5 point 905
For y1 equals 4 point 905, we substitution to x square equals y plus 1, become:
x square equals 4 point 905 plus 1
x square equals 5 point 905
x equals plus minus 2 point 43
For y2 equals minus 5 point 905, we substitution to x square equals y plus 1, become:
x square equals minus 5 point 905 plus 1
x square equals minus 4 point 905, because x square minus so this not valid.
So, point of intersection are 2 point 905 comma 4 point 905 and minus 2 point 905 comma 4 point 905.