Question

Question for Samaritans​

Answer 1

Answer:

$\boxed{x=-2,-1}$

Step-by-step explanation:

Q1.Solve the equation

$\left|\begin{array}{ccc}</p><p>x+2&2x+3&3x+4\\2x+3&3x+4&4x+5\\3x+5&5x+8&10x+17\end{array}\right|=0$

C3-> C3-(C2+C1)

$\left|\begin{array}{ccc}</p><p>x+2&2x+3&3x+4-3x-5\\2x+3&3x+4&4x+5-5x-7\\3x+5&5x+8&10x+17-8x-13\end{array}\right|=0$

$\left|\begin{array}{ccc}</p><p>x+2&2x+3&-1\\2x+3&3x+4&-x-2\\3x+5&5x+8&2x+4\end{array}\right|=0$

C1->2C1-C2

$\left|\begin{array}{ccc}</p><p>2x+4-2x-3&2x+3&-1\\4x+6-3x-4&3x+4&-x-2\\6x+10-5x+8&5x+8&2x+4\end{array}\right|=0$

$\left|\begin{array}{ccc}</p><p>1&2x+3&-1\\x+2&3x+4&-x-2\\x+2&5x+8&2x+4\end{array}\right|=0$

C1->C1+C3

$\left|\begin{array}{ccc}</p><p>0&2x+3&-1\\0&3x+4&-x-2\\3x+6&5x+8&2x+4\end{array}\right|=0$

C2->C2+3C3

$\left|\begin{array}{ccc}</p><p>0&2x&-1\\0&-2&-x-2\\3x+6&11x+20&2x+4\end{array}\right|=0$

Now expand the determinant along C1

$(3x + 6)(2x( - x - 2) - 2) = 0 \\ \\ 3x + 6 = 0 \\ \\ x = - 2 \\ \\ or \\ \\ - 2 {x}^{2} - 4x - 2 = 0 \\ \\ {x}^{2} + 2x + 1 = 0 \\ \\ {x}^{2} + x + x + 1 = 0 \\ \\ x(x + 1) + 1(x + 1) = 0 \\ \\ (x + 1)(x + 1) = 0 \\ \\ x = - 1$

Hope it helps you.

Answer 2

Answer:

what we should solve here