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

Answer:

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

Answer:

#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#.