# Find the value of $p$ if quadratic equation is $x^2+px+12 = 0$ and roots are in the ratio $1:2$

$x^2+px+12 = 0$ is the given quadratic equation and the roots of this equation are in the ratio of $1:2$.

###### Step: 1

If $ax^2 + bx + c = 0$ is the standard form quadratic equation and it represents the given equation. If roots are $\alpha$ and $\beta$, the same roots are also roots of the given equation.

Therefore, $\alpha : \beta = 1:2$

$\implies \dfrac{\alpha}{\beta} = \dfrac{1}{2}$

$\implies \beta = 2 \alpha$

It expresses that the value of $\beta$ is twice the value of $\alpha$.

###### Step: 2

The summation of the roots and the product of the roots of the quadratic equation can be expressed in terms of the literal coefficients of the standard form quadratic equation.

$\alpha + \beta = \,-\dfrac{b}{a}$

$\alpha \beta = \dfrac{c}{a}$

Compare the quadratic equation $x^2+px+12 = 0$ with standard form quadratic equation. Therefore $a = 1$, $b = p$ and $c = 12$.

$\alpha + \beta = \,-\dfrac{p}{1} = -p$

$\alpha \beta = \dfrac{12}{1} = 12$

###### Step: 3

According to step $1$, the value of $\beta$ is equal to $2\alpha$. So, eliminate $\beta$ from the product of the roots by substituting its values in terms of $\alpha$.

$\alpha \beta = \alpha (2 \alpha) = 12$

$\implies 2 \alpha^2 = 12$

$\implies \alpha^2 = \dfrac{12}{2}$

$\require{cancel} \implies \alpha^2 = \dfrac{\cancel{12}}{\cancel{2}}$

$\therefore \,\, \alpha^2 = 6$

###### Step: 4

Similarly, eliminate $\beta$ from the summation of the roots of the quadratic equation by substituting its values in terms of $\alpha$.

$\alpha + 2\alpha = -p$

$\implies 3\alpha = -p$

$\implies \alpha = -\dfrac{p}{3}$

Square both sides.

$\implies \alpha^2 = {\Bigg(-\dfrac{p}{3}\Bigg)}^2$

$\implies \alpha^2 = \dfrac{p^2}{9}$

It is derived in the previous step, the value of $\alpha^2$ is $6$.

$\implies 6 = \dfrac{p^2}{9}$

$\implies 6 \times 9 = p^2$

$\implies p^2 = 6 \times 9$

$\implies p = \pm \sqrt{6 \times 9}$

$\therefore \,\, p = \pm 3 \sqrt{6}$

The values of $p$ are $3 \sqrt{6}$ and $-3 \sqrt{6}$.

It is the required solution for this quadratic equation problem in mathematics.

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