Answer
the required values are\[b=3,4\].
Work Step by Step
Compare the equation \[{{x}^{2}}+bx+5\]with standard quadratic equation, we get
\[\begin{align}
& a=1 \\
& b=4 \\
& c=c \\
\end{align}\]
Condition for real root is:
\[\begin{align}
& {{b}^{2}}-4ac=0 \\
& {{b}^{2}}=4ac \\
& {{4}^{2}}=4\left( 1 \right)\left( c \right) \\
& 16=4c
\end{align}\]
So,
\[\begin{align}
& c=\frac{16}{4} \\
& =4
\end{align}\]
So, \[b=4\]
And,
\[\begin{align}
& {{x}^{2}}+4x+4={{x}^{2}}+2x+2x+4 \\
& =x\left( x+2 \right)+2\left( x+2 \right) \\
& =\left( x+2 \right)\left( x+2 \right)
\end{align}\]
The other value of \[b\]is 3.
So,
\[\begin{align}
& {{x}^{2}}+4x+3={{x}^{2}}+1x+3x+3 \\
& =x\left( x+1 \right)+3\left( x+1 \right) \\
& =\left( x+1 \right)\left( x+3 \right)
\end{align}\]
Hence, the required values are\[b=3,4\].
