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

Question

Kennedy Adams · Accepted Answer

#x = 5# and #x = 3#

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#

Abigail Allen · Answer

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

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) :}#

Nathan Axel · Answer

undefined To solve the quadratic equation $100x^2 - 800x + 1500 = 0$, you can use the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = 100$, $b = -800$, and $c = 1500$. Substituting these values into the formula and simplifying will give you the solutions for $x$.