How do you use the quadratic formula to find both solutions to the quadratic equation x^2+3x-4=0?

2 Answers
Jun 29, 2015

By replacing the variable coefficients in the general quadratic formual with the coefficients for the terms in the given equation we can get x=1 or x=-4

Explanation:

For an equation in the form:
color(white)("XXXX")ax^2+bx+c=0
the quadratic formula gives the solutions as
color(white)("XXXX")x= (-b+-sqrt(b^2-4ac))/(2a)

x^2+3x-4 = 0 is in this form with
color(white)("XXXX")a=1
color(white)("XXXX")b=3 and
color(white)("XXXX")c=(-4)

Replacing a, b, and c in the quadratic formula, we get
color(white)("XXXX")x=(-3+-sqrt(3^2-4(1)(-4)))/(2(1))

color(white)("XXXX")x= (-3+-sqrt(9+16))/2

color(white)("XXXX")x=(-3+-5)/2

color(white)("XXXX")x = 2/2 or x= (-8)/2

x=1 or x=-4

Jun 29, 2015

x_1=-4
x_2=1

Explanation:

Having the equation x^2+3x−4=0, we find the Delta_x:
Delta_x=3^2-4*1*(-4)=9+16=25

Now we find x_1 and x_2 with the quadratic formula .

x_"1,2"=(-3+-sqrt(Delta_x))/(2*1)=(-3+-5)/2.

The two solutions are:

x_1=(-3-5)/2=-4

x_2=(-3+5)/2=1.