How do you solve #2x^2 - 4x - 5=0# using the quadratic formula?

1 Answer
Jan 18, 2017

Answer:

#x = 1+-sqrt(14)/2#

Explanation:

#2x^2-4x-5 = 0#

is of the form:

#ax^2+bx+c = 0#

with #a=2#, #b=-4# and #c=-5#.

This has roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (4+-sqrt((-4)^2-4(2)(-5)))/(2(2))#

#color(white)(x) = (4+-sqrt(16+40))/4#

#color(white)(x) = (4+-sqrt(56))/4#

#color(white)(x) = (4+-sqrt(2^2*14))/4#

#color(white)(x) = (4+-2sqrt(14))/4#

#color(white)(x) = 1+-sqrt(14)/2#

#color(white)()#
Footnote

The quadratic formula is very useful, but is it just a "magical" formula to you, or do you know how to derive it?

Here's one way:

Given:

#ax^2+bx+c = 0#

We find:

#0 = 1/a(ax^2+bx+c)#

#color(white)(0) = x^2+b/ax+c/a#

#color(white)(0) = x^2+2b/(2a)x+b^2/(2a)^2-b^2/(2a)^2+c/a#

#color(white)(0) = (x+b/(2a))^2-(b^2-4ac)/(4a^2)#

Add #(b^2-4ac)/(4a^2)# to both ends and transpose to get:

#(x+b/(2a))^2 = (b^2-4ac)/(4a^2)#

Take the square root of both sides, allowing for both positive and negative square roots to find:

#x+b/(2a) = +-sqrt((b^2-4ac)/(4a^2))= +-sqrt(b^2-4ac)/(2a)#

Subtract #b/(2a)# from both ends to find:

#x = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b+-sqrt(b^2-4ac))/(2a)#