
MHB Master
#1
October 2nd, 2017,
01:28
what is a b c and d so that all values of s are true
\begin{align}\displaystyle
&f_{15}=\\
&17d+11s^24s+10as^3=(b+2)s+90s^3+(3c1)s^2+85\\
&17d+11s^26s+10as^3=bs+90s^3+3cs^2s^2+85\\
&(s=0)\\
&17d=85 \therefore d=5\\
&11s^26s+10as^3=bs+90s^3+3cs^2s^2\\
&(s=1)\\
&11+610a=b90+3c1\\
&10a+b3c=18 \\
&(s=1) \\
&116+10a=b+90+3c1\\
&10ab3c=6 \\
&3c=24\therefore c=4
\end{align}
what i couldn't get is a and b

October 2nd, 2017 01:28
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Pessimist Singularitarian
#2
October 2nd, 2017,
01:42
From the first equation, we have be equating like coefficients:
$ \displaystyle 10a=90$
$ \displaystyle 11=3c1$
$ \displaystyle 4=b+2$
$ \displaystyle 17d=85$
Solve each of these to find the values in question.

#3
October 2nd, 2017,
09:39
Or: you have four unknowns, a, b, c, and d so you need four equations, not just three. Take one more values for s, say s= 2, to get one more equation.

MHB Master
#4
October 2nd, 2017,
13:36
Thread Author
mahalo much
I tried earlier to do this by some factoring
and a thot a matrix could be used
but ran into fog banks

Pessimist Singularitarian
#5
October 2nd, 2017,
13:50
Originally Posted by
karush
mahalo much
I tried earlier to do this by some factoring
and a thot a matrix could be used
but ran into fog banks
You could use an augmented matrix:
$ \displaystyle \left[\begin{array}{ccccc}10 & 0 & 0 & 0 & 90 \\ 0 & 1 & 0 & 0 & 6 \\ 0 & 0 & 3 & 0 & 12 \\ 0 & 0 & 0 & 17 & 85 \\ \end{array}\right]$
Now perform:
$ \displaystyle \frac{1}{10}R_1,\,\frac{1}{3}R_3,\,\frac{1}{17}R_4$
to obtain:
$ \displaystyle \left[\begin{array}{ccccc}1 & 0 & 0 & 0 & 9 \\ 0 & 1 & 0 & 0 & 6 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \\ \end{array}\right]$