1st 3 terms of (1+x)^n are 1-0.9+ 0.36. Find 'x' and 'n'

In the expansion of (1+x)^n, the first three terms are 1-0.9+ 0.36. Find the values of x and n.

How would you find n since they only give you the first 3 terms and dont show you the whole expansion. If I can figure out n, i'll prob be able to figure out x.

Thanks again

Maybe it's me, but how is 1-0.9+0.36 three terms?. Your post seems rather baffling. Can you elaborate?.

[1+x]^n=1^n+n1^(n-1)x+n[n-1]1^(n-2)x^2/2
[1+x]^n=1+nx+n[n-1]x^2

given
eq1)
1=1
eq2)
nx=-.9
eq3)
n[n-1]x^2=.36

substitute x=-.9/n into equation 3
then x^2=.81/n^2

n[n-1][.81]=.36n^2
.81n^2-.81n=.36n^2
.45n^2-.81n=0
n[.45n-.81]=0
n=0 or .45n=.81
n=0 or n=9/5
n=0 or n=1.8

n=0 x=0 trivial solution
n=1.8 x=-.9/1.8 or x=-1/2 answer

please check for errors
Arthur

1- 0.9 +0.36

1 term 2 term 3term

n= the amount of terms minus one

answer in the back of book: x= -0.1, n =9

\(\displaystyle \L {n \choose 0} + {n \choose 1} x +{n \choose 2} x^2\)

\(\displaystyle \L 1 = 1\quad ,\quad nx = - 0.9,\quad \& \quad \left( {\frac{{n^2 - n}}{2}} \right)x^2 = .36\)

\(\displaystyle \L \begin{array}{rcl}
\left( {n^2 - n} \right)x^2 & = & .72 \\
nx & = & - 0.9 \\
n^2 x^2 & = & 0.81 \\
- nxx & = & - 0.9 \\
x & = & \frac{1}{{10}}\quad \& \quad n = 9 \\
\end{array}\)