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The roots ${x_1}$ and ${x_2}$ of the equation ${x^2} + px + 12 = 0$, are such that ${x_2} - {x_1} = 1$. Find p.

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Last updated date: 13th Jul 2024
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Answer
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Hint: We will solve the problem using the properties of quadratic equation like: -
Sum of roots of a general quadratic equation is $\dfrac{{ - b}}{a}$ and the product of roots of a general quadratic equation is $\dfrac{c}{a}$.

Complete step by step answer:
Given that: ${x^2} + px + 12 = 0$ and ${x_2} - {x_1} = 1$.
So from the equation ${x^2} + px + 12 = 0$ as we know that if the equation is in the form of $a{x^2} + bx + c = 0$and the roots are ${x_1}\& {x_2}$.
Then the sum of roots $\left( {{x_1} + {x_2}} \right) = \dfrac{{ - b}}{a}$
And the product of roots $\left( {{x_1} \times {x_2}} \right) = \dfrac{c}{a}$
By comparing the given equation with general equation, we have
$a = 1,b = p,c = 12$
Also$\left( {{x_1} + {x_2}} \right) = \dfrac{{ - p}}{1} = - p$ - (1)
And given that$\left( {{x_2} - {x_1}} \right) = 1$ - (2)
Adding equation 2 to equation 1 we get:
\[
   \Rightarrow \left( {{x_2} + {x_1}} \right) - \left( {{x_2} - {x_1}} \right) = - p + 1 \\
   \Rightarrow 2{x_2} = - p + 1 \\
   \Rightarrow {x_2} = \dfrac{{1 - p}}{2} \\
\]
From equation 2 we get:
\[
   \Rightarrow {x_1} = {x_2} - 1 \\
   \Rightarrow {x_1} = \dfrac{{1 - p}}{2} - 1 \\
   \Rightarrow {x_1} = \dfrac{{1 - p - 2}}{2} \\
   \Rightarrow {x_1} = \dfrac{{ - 1 - p}}{2} \\
\]
By using the second property further we get:
\[
   \Rightarrow {x_1} \times {x_2} = \dfrac{c}{a} = 12 \\
   \Rightarrow \left( {\dfrac{{ - 1 - p}}{2}} \right)\left( {\dfrac{{1 - p}}{2}} \right) = 12 \\
   \Rightarrow \left( { - 1 - p} \right)\left( {1 - p} \right) = 12 \times 4 = 48 \\
   \Rightarrow - 1 + {p^2} = 48 \\
   \Rightarrow {p^2} = 49 \\
   \Rightarrow p = \pm 7 \\
\]
Hence, p has 2 values that are +7 and -7.

Note: Whenever solving a quadratic equation, we should always keep both the solution and properties of roots of the quadratic equation that are mentioned above in the hint section to solve the problem easily and fast.