Answer
the required values are\[b=8,16\].
Work Step by Step
It is known that the solution of the quadratic equation \[a{{x}^{2}}+bx+c=0\] is;
\[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
The condition \[{{b}^{2}}-4ac\]gives the information that the solution is real, imaginary and equal
If \[{{b}^{2}}-4ac>0\]the solution is real and distinct.
If \[{{b}^{2}}-4ac=0\]the solution is equal.
If \[{{b}^{2}}-4ac<0\]the solution is imaginary or complex.
Compare the equation \[{{x}^{2}}+bx+5\]with the standard quadratic equation;
\[\begin{align}
& a=1 \\
& b=b \\
& c=15
\end{align}\]
The conditionfor real root is;
\[\begin{align}
& {{b}^{2}}-4ac=0 \\
& {{b}^{2}}=4ac \\
& =4\left( 1 \right)\left( 15 \right) \\
& =60
\end{align}\]
So,
\[\begin{align}
& b=\sqrt{60} \\
& =7.74
\end{align}\]
So, put \[b=8\]in\[{{x}^{2}}+bx+15\];
\[\begin{align}
& {{x}^{2}}+8x+15={{x}^{2}}+3x+5x+15 \\
& =x\left( x+3 \right)+5\left( x+3 \right) \\
& =\left( x+3 \right)\left( x+5 \right)
\end{align}\]
The other possibility is\[b=16\].
So,
\[\begin{align}
& {{x}^{2}}+16x+15={{x}^{2}}+1x+15x+15 \\
& =x\left( x+1 \right)+5\left( x+1 \right) \\
& =\left( x+1 \right)\left( x+5 \right)
\end{align}\]
Hence, the required values are\[b=8,16\].
