How do you solve #x^2 - 4x = 5# using the quadratic formula?

2 Answers
Jun 1, 2017

Answer:

#color(blue)(x=-1 or x=5#

Explanation:

General quadratic equation:

#ax^2+bx+c=0#

Quadratic formula:

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

#x^2-4x=5#

#:.x^2-4x-5=0#

#:.a=1,b=-4,c=-5#

#:.x=(-(-4)+-sqrt((-4)^2-4(1)(-5)))/(2(1))#

#:.x=(4+-sqrt(16+20))/2#

#:.x=(4+-sqrt36)/2#

#:.x=(4+-6)/2#

#:.x=(4+6)/2#

or:

#:.x=(4-6)/2#

#:.x=10/2# or #x=(-2)/2#

#:.color(blue)(x=5# or #color(blue)(x=-1#

substitute #color(blue)(x=-1#

#:.(-1)^2-4(-1)=5#

#:.1+4=5#

#:.color(blue)(color(blue)(5=5#

substitute #color(blue)(x=5#

#:.(5)^2-4(5)=5#

#:.25-20=5#

#:.color(blue)(5=5#

Jun 1, 2017

Answer:

#x=5, x=-1#

Explanation:

Subtract #5# from both sides.

#x^2-4x-5=0#

From this, #a=1, b=-4, c=-5#

The quadratic formula is #(-b+-sqrt(b^2-4ac))/(2a)#

Substituting the values

#(-(-4)+-sqrt((-4)^2-4*1*-5))/(2*1)#

#(4+-sqrt(16+20))/2#

#(4+sqrt(36))/2#

#4/2+sqrt(36)/2, 4/2-sqrt(36)/2#

#2+sqrt(36)/2, 2-sqrt(36)/2#

#2+6/2, 2-6/2#

#2+3, 2-3#

#5, -1#