How do you solve the system of equations #y-4x=x^2-2# and #y+7=6x# algebraically?

1 Answer
Nghi N.
Jul 14, 2016

#5 +- 2sqrt5#

Explanation:

#y1 = x^2 - 4x - 2# (1)
y2 = 6x - 7 (2)
The line y1 intersects the parabola y2 at 2 points whose x-coordinates are the 2 real roots of the quadratic equation
y1 = y2
#x^2 - 4x - 2 = 6x - 7#
#x^2 - 10x + 5 = 0#
#D = d^2 = 100 - 20 = 80 = 5(16)# --> #d = 4sqrt5#
There are 2 intersects whose coordinates are:
#x = -b/(2a) +- d /(2a) = 10/2 +- (4sqrt5)/2 = 5 +- 2sqrt5#

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