Question

if after factoring a quadratic equation I end up with something like this:
N(N-2)=72
How can I find the roots? Please help

Answer 1

n^2-2n+1=73
(n-1)^2=73
n-1= + or - (73)^(1/2)
n=+ or - (73)^(1/2) +1

Answer 2

This method is called completing the square, and works with all quadratic forms.

Answer 3

\[n^2 - 2n = 73\]
\[use\ the\ sridharan's\ formula,\ i.e\ x=1\pm 2\ sqrt{74}\]

Answer 4

Sorry to note, but the above answer did not properly transcribe the original equation, thus yielding an incorrect solution.

Answer 5

x1=1+sqrt73 ans
x2=1-sqrt73 answ

Answer 6

did you get it emunrra?

Answer 7

No really

Answer 8

use the quadratic formula in these eq ax^2 + bx +c
x=(-b-+sqrt(b^2-4ac))/2a

Answer 9

N(N-2)=72 is already factored. You dont have to expand it (that is what i think)
so for me x1 and x2 must be:
x1=72
x2=74
Not sure of that

Answer 10

N(N-2)=72
N^2 -2N =72
N^2 -2N -72=0 USE a=1, b=-2,c=-72

Answer 11

N^2 -2N -72=0 this is also
x^2 -2x -72=0 USE a=1, b=-2,c=-72

Answer 12

You do not need to resort to the formula. You can solve this problem without it, using only fundamental arithmetic operations.
\[N(N-2)=72\]
\[N^2-2N=72\]
\[N^2-2N+1=73\]
\[(N-1)^2=73\]
\[N-1=\pm\sqrt{73}\]
\[N= \pm\sqrt{73}\ +1\]

Answer 13

ahhh ok. Got it.

Answer 14

correct, if you use factoring

Answer 15

thank you very much

Answer 16

yw emunrra...lol..good luck...