How do you solve #100x^2-800x+1500=0#?

2 Answers
Dec 12, 2016

Answer:

#x = 5# and #x = 3#

Explanation:

First, divide both sides of the equation by #100# to make this simpler to work with:

#1/100(100x^2 - 800x + 1500) = 0/100#

#(100x^2)/100 - (800x)/100 + 1500/100 = 0#

#x^2 - 8x + 15 = 0#

Now we can play with multipliers of 15 (1x15, 3x5, 5x3, 15x1) to factor the quadratic equation:

#(x - 5)(x - 3) = 0#

We can now solve each term for #0#:

#x - 5 = 0#

#x - 5 + 5 = 0 + 5#

#x - 0 = 5#

#x = 5#

and

#x - 3 = 0#

#x - 3 + 3 = 0 + 3#

#x - 0 = 3#

#x = 3#

Dec 12, 2016

Answer:

#color(green)(x=3)# or #color(green)(x=5)#

Explanation:

If
#color(white)("XXX")100x^2-800x+1500=0#
then (after dividing both sides by #100#)
#color(white)("XXX")x^2-8+15=0#

We would like to factor this in the form:
#color(white)("XXX")(x-a)(x-b)=0#
and since
#color(white)("XXX")(x-a)(x-b)=x^2+(-a-b)x+ab#
we are looking for values of #a# and #b#
such that
#color(white)("XXX")-a-b=-8 rarr a+b=8#
and
#color(white)("XXX")ab=15#

Checking factors of #15#, we quickly find the pair #3# and #5# that satisfy our requirement.

So we have
#color(white)("XXX")(x-3)(x-5)=0#

From which it follows that
#{: ("either ",(x-3)=0," or ",(x-5)=0), (,rarr x=3,,rarrx=5) :}#