How do you solve #x² - x - 20 = 0#?

1 Answer
George C.
Jul 13, 2015

Find two numbers whose product is #20# and whose difference is #1#. The pair #5#, #4# works.

Hence #x^2-x-20 = (x-5)(x+4)#.

So solutions are #x=5# or #x=-4#.

Explanation:

#(x-a)(x+b) = x^2 - (a-b)x - ab#

Matching this against #x^2-x-20# we see that if we can find #a# and #b# such that #a - b = 1# and #ab = 20#, then we can factor the quadratic into two linear terms.

The pair #a=5#, #b=4# works, giving us

#x^2-x-20 = (x-5)(x+4)#

This will be #0# when #(x-5) = 0# or #(x+4) = 0#, which is when #x=5# or #x=-4#.

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