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Home » If     and     are the roots of the equation     px² + 7x + q = 0, find the values of p and q.
If

    \[2/3\]

and

    \[-3\]

are the roots of the equation

    \[\mathbf{px}{}^\text{2}\text{ }+\text{ }\mathbf{7x}\text{ }+\text{ }\mathbf{q}\text{ }=\text{ }\mathbf{0}\]

px² + 7x + q = 0, find the values of p and q.
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Quadratic Equations in One Variable
If

    \[2/3\]

and

    \[-3\]

are the roots of the equation

    \[\mathbf{px}{}^\text{2}\text{ }+\text{ }\mathbf{7x}\text{ }+\text{ }\mathbf{q}\text{ }=\text{ }\mathbf{0}\]

px² + 7x + q = 0, find the values of p and q.

Let us substitute the given value x =

    \[2/3\]

in the expression, we get

    \[\begin{array}{*{35}{l}} px{}^\text{2}\text{ }+\text{ }7x\text{ }+\text{ }q\text{ }=\text{ }0 \\ p{{\left( 2/3 \right)}^{2}}~+\text{ }7\left( 2/3 \right)\text{ }+\text{ }q\text{ }=\text{ }0 \\ 4p/9\text{ }+\text{ }14/3\text{ }+\text{ }q\text{ }=\text{ }0 \\ By\text{ }taking\text{ }LCM \\ 4p\text{ }+\text{ }42\text{ }+\text{ }9q\text{ }=\text{ }0 \\ \end{array}\]

    \[4p\text{ }+\text{ }9q\text{ }=\text{ }\text{ }42\]

… (1)

Now, substitute the value x =

    \[-3\]

in the expression, we get

    \[\begin{array}{*{35}{l}} px{}^\text{2}\text{ }+\text{ }7x\text{ }+\text{ }q\text{ }=\text{ }0 \\ p{{\left( -3 \right)}^{2}}~+\text{ }7\left( -3 \right)\text{ }+\text{ }q\text{ }=\text{ }0 \\ 9p\text{ }+\text{ }q\text{ }\text{ }21\text{ }=\text{ }0 \\ 9p\text{ }+\text{ }q\text{ }=\text{ }21 \\ \end{array}\]

    \[q\text{ }=\text{ }21\text{ }\text{ }9p\]

…. (2)

By substituting the value of q in equation (1), we get

    \[\begin{array}{*{35}{l}} 4p\text{ }+\text{ }9q\text{ }=\text{ }\text{ }42 \\ 4p\text{ }+\text{ }9\left( 21\text{ }\text{ }9p \right)\text{ }=\text{ }-42 \\ 4p\text{ }+\text{ }189\text{ }\text{ }81p\text{ }=\text{ }-42 \\ 189\text{ }\text{ }77p\text{ }=\text{ }-42 \\ 189\text{ }+\text{ }42\text{ }=\text{ }77p \\ 231\text{ }=\text{ }77p \\ p\text{ }=\text{ }231/77 \\ p\text{ }=\text{ }3 \\ \end{array}\]

Now, substitute the value of p in equation (2), we get

    \[\begin{array}{*{35}{l}} q\text{ }=\text{ }21\text{ }\text{ }9p \\ =\text{ }21\text{ }\text{ }9\left( 3 \right) \\ =\text{ }21\text{ }\text{ }27 \\ =\text{ }-6 \\ \end{array}\]

∴ Value of p is

    \[3\]

and q is

    \[-6\]

.

i

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