How do you solve #2x^2+4x-5=0#?

1 Answer
George C.
Mar 25, 2016

Use the quadratic formula to find:

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

Explanation:

#2x^2+4x-5# is in the form #ax^2+bx+c# with #a=2#, #b=4# and #c=-5#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 4^2-(4*2*(-5)) = 16+40 = 56 = 2^2*14#

This is positive, but not a perfect square. So our quadratic equation has Real irrational roots.

We can find the roots using the quadratic formula:

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

#=(-b+-sqrt(Delta))/(2a)#

#=(-4+-sqrt(56))/(2*2)#

#=(-4+-2sqrt(14))/4#

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

#color(white)()#
Alternative Method

Multiply through by #2# first to make the leading term a perfect square, complete the squre, then factorise using the difference of square identity:

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

with #a=(2x+2)# and #b=sqrt(14)# as follows:

#0 = 4x^2+8x-10#

#=(2x+2)^2-4-10#

#=(2x+2)^2-(sqrt(14))^2#

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

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

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

#=4(x+1-sqrt(14)/2)(x+1+sqrt(14)/2)#

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

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