Question

[tex] \left |\begin{array}{ccc} x&5&9 \\ 16&3x+8&36 \\ 3& 1&7\end{array}\right | =0,Find the Solution set. [\tex]

Answer 1

Step-by-step explanation:

$\left |\begin{array}{ccc} x&5&9 \\ 16&3x+8&36 \\ 3& 1&7\end{array}\right | =0$

For simplification ,R1->R1-5R3

$\left |\begin{array}{ccc} x-15&5-5&9-35 \\ 16&3x+8&36 \\ 3& 1&7\end{array}\right | =0$

$\left |\begin{array}{ccc} x-15&0&-26\\ 16&3x+8&36 \\ 3& 1&7\end{array}\right | =0$

C3->C3-7C2

$\left |\begin{array}{ccc} x-15&0&-26-0\\ 16&3x+8&36-21x-56 \\ 3& 1&7-7\end{array}\right | =0$

$\left |\begin{array}{ccc} x-15&0&-26\\ 16&3x+8&-21x-20 \\ 3& 1&0\end{array}\right | =0$

expand the determinant along R1

$(x - 15)(+21x+20) - 26(16 - 9x - 24) = 0 \\ \\ 21 {x}^{2} + 20x - 315x - 300 - 26(-8 - 9x) = 0 \\ \\ 21 {x}^{2} + 20x - 315x - 300 +208+ 234x= 0 \\ \\ 21 {x}^{2} - 61x - 92 = 0 \\ \\ x_{1,2} = \frac{ 61± \sqrt{4825} }{42} \\ \\ x_{1,2} = \frac{ 61± 5\sqrt{193} }{42} \\ \\ x_{1}=\frac{ 61+ 5\sqrt{193} }{42}\\\\x_{2}=\frac{ 61- 5\sqrt{193} }{42}$

Hope it helps you