How do you solve #(42- x ) ( 25- x ) = 500#?

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Answer 1 · 2018-07-25T16:26:03

#x = (67 pm sqrt 2289)/2#

#42*25 - 42x - 25x + x^2 - 500 = 0#

#x^2 - 67x + 550 = 0#

#Delta = 67^2 - 4 * 550 = 4489 - 2200 = 2289#

Answer 2 · 2018-07-25T16:28:02

#x=\frac{67\pm\sqrt2289}{2}#

Given that

#(42-x)(25-x)=500#

#x^2-67x+1050=500#

#x^2-67x+550=0#

#x=\frac{-(-67)\pm\sqrt{(-67)^2-4(1)(550)}}{2(1)}#

#x=\frac{67\pm\sqrt2289}{2}#

Answer 3 · 2018-07-25T16:44:59

#x=(67+sqrt2289)/2 or x=(67-sqrt2289)/2#

Here,

#(42-x)(25-x)=500#

Simplifying we get ,

#42(25-x)-x(25-x)=500#

#1050-42x-25x+x^2=500#

#x^2-67x+1050-500=0#

#x^2-67x+550=0#

Comparing with #ax^2+bx+c=0# ,we get

#a=1 ,b=-67 and c=550#

#Delta=b^2-4ac=(-67)^2-4(1)(550)#

#Delta=4489-2200=2289#

So,

#x=(-b+-sqrtDelta)/(2a)=(67+-sqrt2289)/2#

#:.x=(67+sqrt2289)/2 or x=(67-sqrt2289)/2#

#:.x~~57.4217 or x~~9.5783#