Find the values of x if \[\mathbf{p}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]and \[{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\].

Find the values of x if

$\mathbf{p}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$

and

${{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}$

Find the values of x if

$\mathbf{p}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$

and

${{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}$

Given quadratic equation:

${{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}$

And,

$\mathbf{p}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$

So,

$p\text{ }=\text{ }-1$

Substituting the value of p in the given quadratic equation, we get

$\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }\left( -1 \right)x\text{ }\text{ }6\text{ }=\text{ }0 \\ {{x}^{2}}~\text{ }x\text{ }\text{ }6\text{ }=\text{ }0 \\ \end{array}$

Solving for x by factorization, we have

$\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }3x\text{ }+\text{ }2x\text{ }\text{ }6\text{ }=\text{ }0 \\ x\left( x\text{ }\text{ }3 \right)\text{ }+\text{ }2\left( x\text{ }\text{ }3 \right)\text{ }=\text{ }0 \\ \left( x\text{ }+\text{ }2 \right)\text{ }\left( x\text{ }\text{ }3 \right)\text{ }=\text{ }0 \\ \end{array}$

So,

$\begin{array}{*{35}{l}} x\text{ }+\text{ }2\text{ }=\text{ }0\text{ }or\text{ }x\text{ }\text{ }3\text{ }=\text{ }0 \\ x\text{ }=\text{ }-2\text{ }or\text{ }x\text{ }=\text{ }3 \\ \end{array}$

∴ Value of x =

$-2,\text{ }3$

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