Question

If the absolute difference between two roots of the equation ${x^2} + px + 3 = 0$ is $\sqrt p $ then $p$ is equals to:
A. $ - 3,4$
B. $4$
C. $ - 3$
D. None of these

Answer 1

Hint: Assume that ${x_1}$ and ${x_2}$ are the two roots of the given quadratic equation. Write sum of roots and product of roots (i.e. if we have quadratic equation $a{x^2} + bx + c = 0$ then sum of roots is $\dfrac{{ - b}}{a}$ and product of roots is $\dfrac{c}{a}$ ). Then by using sum and product of roots find the difference of roots. Equate the absolute value of difference of roots with $\sqrt p $. Solve the equation to get the value of $p$ .

Complete step-by-step answer:
If $\alpha $ and $\beta $ are roots of quadratic equation $a{x^2} + bx + c = 0$ then sum of roots (i.e.$\left( {\alpha + \beta } \right)$) is $\dfrac{{ - b}}{a}$ and product of roots (i.e.$\alpha \beta $) is$\dfrac{c}{a}$.
Now let ${x_1}$ and ${x_2}$be the roots of given quadratic equation ${x^2} + px + 3 = 0$
Then sum of roots $ = {x_1} + {x_2} = \dfrac{{ - p}}{1}$ and product of roots $ = {x_1}{x_2} = \dfrac{3}{1}$
$ \Rightarrow {x_1} + {x_2} = - p$ and \[{x_1}{x_2} = 3\]
By using the values of ${x_1} + {x_2}$ and ${x_1}{x_2}$ we can find the value of ${x_1} - {x_2}$
(Example: suppose there are two numbers $a$ and $b$. If their sum and product is given then by using the identity ${\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab$ we can find the difference of two numbers.)
Use the above mentioned identity on ${x_1}$ and ${x_2}$
 $ \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}.....\left( 1 \right)$
Substitute the values of ${x_1} + {x_2}$ and ${x_1}{x_2}$ in the equation $\left( 1 \right)$
$ \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = {\left( { - p} \right)^2} - \left( 4 \right)\left( 3 \right)$
$ \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = {p^2} - 12$
Take square root on both sides
$ \Rightarrow {x_1} - {x_2} = \pm \sqrt {{p^2} - 12} ....\left( 2 \right)$
Given that absolute value of difference of roots is $\sqrt p $
$ \Rightarrow \left| {{x_1} - {x_2}} \right| = \sqrt p ....\left( 3 \right)$
Squaring on both sides
$ \Rightarrow {\left| {{x_1} - {x_2}} \right|^2} = {\left( {\sqrt p } \right)^2}$
$ \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = p$ $...(4)$
Substitute the value of ${x_1} - {x_2}$ in $(4)$ from $(2)$
$ \Rightarrow {\left( { \pm \sqrt {{p^2} - 12} } \right)^2} = p$
$ \Rightarrow {p^2} - 12 = p$
$ \Rightarrow {p^2} - p - 12 = 0$ this is the quadratic in $p$.
Roots of the quadratic equation $a{x^2} + bx + c = 0$ by $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Apply the above mentioned formula to find the roots of quadratic equation ${p^2} - p - 12 = 0$
$ \Rightarrow p = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 12} \right)} }}{{2 \times 1}}$
$ \Rightarrow p = \dfrac{{1 \pm \sqrt {1 - 4\left( { - 12} \right)} }}{2}$
On simplifying we get $p = \dfrac{{1 \pm \sqrt {1 + \left( {48} \right)} }}{2} \Rightarrow p = \dfrac{{1 \pm \sqrt {49} }}{2}$
$ \Rightarrow p = \dfrac{{1 \pm 7}}{2}$
$ \Rightarrow p = \dfrac{{1 + 7}}{2}$ or $p = \dfrac{{1 - 7}}{2}$
+$ \Rightarrow p = \dfrac{8}{2}$ or $p = \dfrac{{ - 6}}{2}$
$ \Rightarrow p = - 3$ or $4$
We can’t take $p = - 3$ because if we put $p = - 3$ in equation $\left( 3 \right)$ then the square root of $p$ is not defined.
So $p = 4$
Hence Option (B) is the correct answer.

Note: If you forgot the formula of sum of roots and product of roots in exam. Then find the roots of the given equation by discriminant formula (Roots of the quadratic equation $a{x^2} + bx + c = 0$ are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$) then find the sum and product of roots. After that follow the same steps done in the solution.