How do you solve #2(6 - x) = x(x + 5) # using the quadratic formula?

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Answer 1 · 2016-06-06T14:39:22

#x=(-7+-sqrt(97))/2#

First let's operate to have the standard form:
#12-2x=x^2+5x#
#x^2+7x-12=0#

#x=(-7+-sqrt(7^2-4*(-12)))/2#
#x=(-7+-sqrt(49+48))/2#
#x=(-7+-sqrt(97))/2#

Answer 2 · 2016-06-06T15:08:58

#x' = (-7 + sqrt(97))/2 #

#x' = (-7- sqrt(97))/2#

#2(6-x)=x(x+5)#

#12-2x=x^2+5x#

#x^2+5x+2x-12=0#

#x^2+7x-12=0#

Your quadratic equation of the form

#ax^2+bx+c=0#

The discriminant is determined using the following formula:

#color(red)(Delta = b^2- 4ac)#

#a=1 \ ; \ b= 7 ; \ c=12#

#Delta = (7)^2 - 4xx1xx(-12)#

#Delta = 97#

Since #Delta>0 => # two real roots exist

# color(red)(x'= (-b + sqrt(Delta))/(2a)) " and " color(red)( x''= (-b - sqrt(Delta))/(2a))#

#x' = (-17 + sqrt(97))/2 ~=1.42#

#x''= (-17 - sqrt(97))/2~=-8.42#