How do you use the quadratic formula to solve #1/4x^2+5x-4=0#?

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Answer 1 · 2018-06-05T17:54:27

#x=-10pm2sqrt(29)#

Multiplying by #4# we get
#x^2+20x-16=0#
so we have
#x_{1,2}=-10pm2\sqrt(29)#

Answer 2 · 2018-06-05T18:04:57

#x= 0.770 or x = -20.77#

You can multiply each term in the equation by #4# to get rid of the fraction. This does not change the equation or the solutions.

#1/4x^2 + 5x - 4=0#

#color(blue)(4)xx 1/4x^2 + color(blue)(4)xx5x- color(blue)(4) xx4=color(blue)(4)xx0#

#x^2+20x-16=0#

This is now in the form #ax^2 +bx+c=0#

#a =1, b=20 and c=-16#

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

#x = (-(20)+-sqrt((20)^2-4(1)(-16)))/(2(1))#

#x = (-20+-sqrt(400+64))/2#

#x = (-20+-sqrt464)/2#

#x= 0.770 or x = -20.77#

Answer 3 · 2018-06-05T18:10:47

#x=-10+-2sqrt(29)#

#1/4x^2+5x-4=0#

First let's get rid of the fraction, we can put a fraction in the quadratic formula but it is easier not to:

#4(1/4x^2+5x-4=0)#

#x^2+20x-16=0#

#y=ax^2 + bx +c#

a = 1

b= 20

c=-16

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

#x=(-20+-sqrt(20^2-4*1*(-16)))/(2*1)#

#x=(-20+-sqrt(400+64))/2#

#x=(-20+-4sqrt(29))/2#

#x=-10+-2sqrt(29)#