How do you solve #x^2+4x-3=0# using the quadratic formula?

1 Answer
Apr 19, 2016

Answer:

Use the quadratic formula to find:

#x = -2+-sqrt(7)#

Explanation:

#x^2+4x-3=0# is of the form #ax^2+bx+c=0#, with #a=1#, #b=4# and #c=-3#.

This has roots given by the quadratic formula:

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

#=(-4+-sqrt(4^2-(4*1*(-3))))/(2*1)#

#=(-4+-sqrt(16+12))/2#

#=(-4+-sqrt(28))/2#

#=(-4+-sqrt(2^2*7))/2#

#=(-4+-2sqrt(7))/2#

#=-2+-sqrt(7)#

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Using the quadratic formula is similar in some ways to completing the square:

#0 = x^2+4x-3#

#=x^2+4x+4-7#

#=(x+2)^2-(sqrt(7))^2#

#=((x+2)-sqrt(7))((x+2)+sqrt(7))#

#=(x+2-sqrt(7))(x+2+sqrt(7))#

Hence #x = -2+-sqrt(7)#

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In fact, we can derive the quadratic formula as follows:

Given #ax^2+bx+c = 0#

#0 = ax^2+bx+c#

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

Hence:

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

Divide both sides by #a# to get:

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

Hence:

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

So:

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