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

3 Answers
Sonnhard
Jun 5, 2018

x=-10pm2sqrt(29)

Explanation:

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

EZ as pi
Jun 5, 2018

x= 0.770 or x = -20.77

Explanation:

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

Bill Jorgensen
Jun 5, 2018

x=-10+-2sqrt(29)

Explanation:

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)

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