Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: (i) \[\mathbf{px}{}^\text{2}\text{ }\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}\] (ii) \[\mathbf{x}{}^\text{2}\text{ }+\text{ }\left( \mathbf{p}\text{ }\text{ }\mathbf{3} \right)\mathbf{x}\text{ }+\text{ }\mathbf{p}\text{ }=\text{ }\mathbf{0}\]

Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: (i)

$\mathbf{px}{}^\text{2}\text{ }\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}$

(ii)

$\mathbf{x}{}^\text{2}\text{ }+\text{ }\left( \mathbf{p}\text{ }\text{ }\mathbf{3} \right)\mathbf{x}\text{ }+\text{ }\mathbf{p}\text{ }=\text{ }\mathbf{0}$

Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: (i)

$\mathbf{px}{}^\text{2}\text{ }\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}$

(ii)

$\mathbf{x}{}^\text{2}\text{ }+\text{ }\left( \mathbf{p}\text{ }\text{ }\mathbf{3} \right)\mathbf{x}\text{ }+\text{ }\mathbf{p}\text{ }=\text{ }\mathbf{0}$

(i)

$\mathbf{px}{}^\text{2}\text{ }\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}$

Let us consider,

$a\text{ }=\text{ }p,\text{ }b\text{ }=\text{ }-4,\text{ }c\text{ }=\text{ }3$

By using the formula,

$\begin{array}{*{35}{l}} D\text{ }=\text{ }{{b}^{2}}~\text{ }4ac \\ =\text{ }{{\left( -4 \right)}^{2}}~\text{ }4\left( p \right)\text{ }\left( 3 \right) \\ =\text{ }16\text{ }\text{ }12p \\ \end{array}$

Since, roots are real.

$\begin{array}{*{35}{l}} 16\text{ }\text{ }12p\text{ }=\text{ }0 \\ 16\text{ }=\text{ }12p \\ p\text{ }=\text{ }16/12 \\ =\text{ }4/3 \\ \therefore p\text{ }=\text{ }4/3 \\ \end{array}$

(ii)

$\mathbf{x}{}^\text{2}\text{ }+\text{ }\left( \mathbf{p}\text{ }\text{ }\mathbf{3} \right)\mathbf{x}\text{ }+\text{ }\mathbf{p}\text{ }=\text{ }\mathbf{0}$

Let us consider,

$a\text{ }=\text{ }1,\text{ }b\text{ }=\text{ }\left( p\text{ }\text{ }3 \right),\text{ }c\text{ }=\text{ }p$

By using the formula,

$\begin{array}{*{35}{l}} D\text{ }=\text{ }{{b}^{2}}~\text{ }4ac \\ =\text{ }{{\left( p\text{ }\text{ }3 \right)}^{2}}~\text{ }4\left( 1 \right)\text{ }\left( p \right) \\ =\text{ }{{p}^{2}}~\text{ }{{3}^{2}}~\text{ }2\left( 3 \right)\text{ }\left( p \right)\text{ }\text{ }4p \\ =\text{ }{{p}^{2}}~+\text{ }9\text{ }\text{ }6p\text{ }\text{ }4p \\ =\text{ }{{p}^{2}}~\text{ }10p\text{ }+\text{ }9 \\ \end{array}$

Since, roots are real and have equal roots.

${{p}^{2}}~\text{ }10p\text{ }+\text{ }9\text{ }=\text{ }0$

Now let us factorize,

$\begin{array}{*{35}{l}} {{p}^{2}}~\text{ }9p\text{ }\text{ }p\text{ }+\text{ }9\text{ }=\text{ }0 \\ p\left( p\text{ }\text{ }9 \right)\text{ }\text{ }1\text{ }\left( p\text{ }\text{ }9 \right)\text{ }=\text{ }0 \\ \left( p\text{ }\text{ }9 \right)\text{ }\left( p\text{ }\text{ }1 \right)\text{ }=\text{ }0 \\ So, \\ \left( p\text{ }\text{ }9 \right)\text{ }=\text{ }0\text{ }or\text{ }\left( p\text{ }\text{ }1 \right)\text{ }=\text{ }0 \\ p\text{ }=\text{ }9\text{ }or\text{ }p\text{ }=\text{ }1 \\ \therefore p\text{ }=\text{ }1,\text{ }9 \\ \end{array}$

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