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    Given that one root of equation $x^2+px +12=0$ is 4 find the value of p and other root.
    ### Step 1: Finding the value of \(p\) using the root \(x = 4\)

    Given \(x^2 + px + 12 = 0\) and one root \(x = 4\), substituting \(x = 4\) gives us:

    \[
    4^2 + 4p + 12 = 0
    \]
    \[
    16 + 4p + 12 = 0
    \]
    \[
    28 + 4p = 0
    \]
    \[
    4p = -28
    \]
    \[
    p = -7
    \]

    ### Step 2: Finding the other root

    With \(p = -7\), our equation becomes \(x^2 – 7x + 12 = 0\).

    Given that the sum of the roots (\(x_1 + x_2\)) of a quadratic equation \(ax^2 + bx + c = 0\) is \(-\frac{b}{a}\), and the product (\(x_1 \cdot x_2\)) is \(\frac{c}{a}\), we can find the other root without solving the equation again.

    Since one root is \(4\), and we know \(p = -7\), we can find the other root by:

    Sum of the roots = \(x_1 + x_2 = -\frac{b}{a} = -\frac{-7}{1} = 7\)

    Thus, \(4 + x_2 = 7\)

    So, \(x_2 = 3\).

    Therefore, with one root being \(4\), and the value of \(p\) is \(-7\), the other root of the equation is \(3\).

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