How do you solve #(42- 2x ) ( 25- 2x ) = 500#?

1 Answer
Jun 13, 2017

Taken to a point where you can finish it.

Explanation:

Lets make the numbers less. Divide both sides by 2

#1/2(42-2x)(25-2x)=500/2#

#color(blue)((21-x))color(green)((25-2x))=250#

Multiply everything in the right brackets by everything in the left.

#color(green)(color(blue)(21)(25-2x)color(blue)(-x)(25-2x)=250#

#525-42x-25x+2x^2=250#

#2x^2-67x+275=0#

Compare to #y=0=ax^2+bx+c#

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

Where #a=2"; "b=-67"; "c=275#

Thus we have:

#x=(+67+-sqrt((-67)^2-4(2)(275)))/(2(2))#

#color(brown)("I will let you finish this off.")#

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Note that as #2x^2# is positive the graph is of form #uu#
Thus the vertex is a minimum.

#x_("vertex") =(-1/2)xxb/a" "->" "(-1/2)xx(-67)/2 = +67/4#

#y_("vertex")=2(67/4)^2-67(67/4)+275 = -286 1/8#

As #" "y_("vertex")<0# and the graph is of form #uu# then solutions exists to #y=0=2x^2-67x+275#