Question

Find those values of c for which the equations
$2x+3y=3$
$(c+2)x+(c+4)y=(c+6)$
$(c+2{)}^{2}x+(c+4{)}^{2}y=(c+6{)}^2$
are consistent.
Find the sum of squares of all the values of c?

Answer 1

Given system of equations is consistent

$D=0$

$\Rightarrow \left|\begin{array}{ccc}2& 3& 3\\ c+2& c+4& c+6\\ {(c+2)}^2& {(c+4)}^2& {(c+6)}^2\end{array}\right|=0$
$C_1\to C_1-C_3;C_2\to C_2-C_3$
$\left|\begin{array}{ccc}-1& 0& 3\\ -4& -2& c+6\\ -4(2c+8)& -2(2c+10)& {(c+6)}^2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& 0& 3\\ 4& 1& c+6\\ 8c+32& 2c+10& {(c+6)}^2\end{array}\right|=0$
$C_1\to C_1-4C_2$
$\Rightarrow \left|\begin{array}{ccc}1& 0& 3\\ 0& 1& c+6\\ -8& 2c+10& {(c+6)}^2\end{array}\right|=0$
$C_3\to C_3-3C_1$
$\Rightarrow \left|\begin{array}{ccc}1& 0& 0\\ 0& 1& c+6\\ -8& 2c+10& {(c+6)}^2+24\end{array}\right|=0$
$\Rightarrow -c^2-10c=0$

$\Rightarrow c=0,c=-10$

Sum of squares of c is $0+100=100$