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If x = 3 is a solution of the equation

    \[\left( \mathbf{k}\text{ }+\text{ }\mathbf{2} \right){{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{kx}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]

, find the value of k. Hence, find the other root of the equation.
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Quadratic Equations in One Variable
If x = 3 is a solution of the equation

    \[\left( \mathbf{k}\text{ }+\text{ }\mathbf{2} \right){{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{kx}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]

, find the value of k. Hence, find the other root of the equation.

Given equation:

    \[\left( \mathbf{k}\text{ }+\text{ }\mathbf{2} \right){{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{kx}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]

And x = 3 is a solution of the equation

So, upon substituting x = 3 it must satisfy the equation

    \[\begin{array}{*{35}{l}} \left( k\text{ }+\text{ }2 \right){{\left( 3 \right)}^{2}}~\text{ }k\left( 3 \right)\text{ }+\text{ }6\text{ }=\text{ }0 \\ \left( k\text{ }+\text{ }2 \right)\left( 9 \right)\text{ }\text{ }3k\text{ }+\text{ }6\text{ }=\text{ }0 \\ 9k\text{ }+\text{ }18\text{ }\text{ }3k\text{ }+\text{ }6\text{ }=\text{ }0 \\ 6k\text{ }+\text{ }24\text{ }=\text{ }0 \\ 6\left( k\text{ }+\text{ }4 \right)\text{ }=\text{ }0 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} k\text{ }+\text{ }4\text{ }=\text{ }0 \\ k\text{ }=\text{ }-4 \\ \end{array}\]

Now, putting k =

    \[-4\]

in the given equation, we have

    \[\begin{array}{*{35}{l}} \left( -4\text{ }+\text{ }2 \right){{x}^{2}}~\text{ }\left( -4 \right)x\text{ }+\text{ }6\text{ }=\text{ }0 \\ -2{{x}^{2}}~+\text{ }4x\text{ }+\text{ }6\text{ }=\text{ }0 \\ \end{array}\]

    \[{{x}^{2}}~\text{ }2x\text{ }\text{ }3\text{ }=\text{ }0\]

[Dividing by

    \[-2\]

on both sides]

Factorizing the above expression, we get

    \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }3x\text{ }+\text{ }x\text{ }\text{ }3\text{ }=\text{ }0 \\ x\left( x\text{ }\text{ }3 \right)\text{ }+\text{ }1\left( x\text{ }\text{ }3 \right)\text{ }=\text{ }0 \\ \left( x\text{ }+\text{ }1 \right)\left( x\text{ }\text{ }3 \right)\text{ }=\text{ }0 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} x\text{ }+\text{ }1\text{ }=\text{ }0\text{ }or\text{ }x\text{ }\text{ }3\text{ }=\text{ }0 \\ x\text{ }=\text{ }-1\text{ }or\text{ }x\text{ }=\text{ }3 \\ \end{array}\]

Hence, the other root of the given equation is

    \[-1\]

.

i

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