How do you solve #y= x^2-4# and #y= x-2#?

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Answer 1 · 2015-08-10T09:33:00

The answer is #x = -1# ; #x = 2#

Since

#{(y = x^2 -4), (y = x - 2) :}#

we shall have

#x^2 -4 = x - 2#

#x^2 -4 - x + 2 = 0#

It becomes a quadratic equation

#x^2 - x - 2 = 0#

#x^2 - 2x + x - 2 = 0#

#x(x - 2) + 1(x - 2) = 0#

#(x + 1) (x - 2) = 0#

This means that you have

#x + 1 = 0 implies x = -1#

or

#x - 2 = 0 implies x = 2#

If you substitute the two values of #x# in the original equations, they satisfy them simultaneously.