Question-178870

Question Number 178870 by ARUNG_Brandon_MBU last updated on 22/Oct/22

Answered by Rasheed.Sindhi last updated on 22/Oct/22

Rule :

\begin{array}{ccc} \begin{matrix} a \end{matrix} & \begin{matrix} b \end{matrix} \\ \begin{matrix} \frac{a d + b c}{2} \end{matrix} \\ \begin{matrix} c \end{matrix} & \begin{matrix} d \end{matrix} \end{array}

Applying rule :

\begin{array}{ccc} \begin{matrix} x + 2 \end{matrix} & \begin{matrix} x + 1 \end{matrix} \\ \begin{matrix} \frac{(x + 2) (3) + (x) (x + 1)}{2} = 1 \end{matrix} \\ \begin{matrix} x \end{matrix} & \begin{matrix} 3 \end{matrix} \end{array}

\frac{(x + 2) (3) + (x) (x + 1)}{2} = 1

3 x + 6 + x^{2} + x = 2

x^{2} + 4 x + 4 = 0

{(x + 2)}^{2} = 0

x = - 2

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

Great! Thanks Sir