Question 777529
121=100*(1+x)^2
121=100*(1+x)(1+x)
this is where you went wrong. you have to do parenthesis first.
121=100*(x^2+2x+1)
121=(100x+100)(x+1)
121=100x^2+100x+100x+100
121=100x^2+200x+100
-121___________-121
0=100x^2+200x-21
you have to then multiply 100 and -21.
you get -2100. you then have to find two numbers that when added equal 200 and when multiplied equal -2100. the numbers are 210 and -10.
the you put them in the equation like so.
0=100x^2+210x-10x-21
then you factor by grouping.
you get:0=10x(10x+21)-1(10x+21)
if you did it correctly you should get the same number in the parenthesis.
then you ass the 10x and the -1 together
0=(10x+21)(10x-1)
then you separate the problem into: 10x+21=0 and 10x-1=0 and solve them
10x+21=0
subtract 21 to 0
10x=-21
divide 10 to -21
x=-21/10
but your not done because you still have to do 10x-1=0
10x-1=0
add 1 to 0
10x=1
divide 10 to 1
x=1/10
____
the two answers are x=1/10, and -21/10