Question

One of the roots of ${x^2} + 3x + 2 = 0$ and ${x^2} + 4x + m = 0$ is common. Value of is/are: -
A) 4
B) 3
C) 1
D) 2

Answer 1

Hint: Here in this question we will find the roots of both equations and then will compare the equal roots to find the value of m. To find the roots of quadratic equations we will use formula which is mentioned below:-
${x_1},{x_2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ Where ${x_1},{x_2}$ are the two roots of quadratic equation.

Complete step-by-step answer:
First of all we will find the roots of the quadratic equation ${x^2} + 3x + 2 = 0$ by applying the formula.
Comparing this equation ${x^2} + 3x + 2 = 0$with the general equation $a{x^2} + bx + c = 0$
Here a=1, b=3 and c=2
$x = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4 \times 1 \times 2} }}{{2 \times 1}}$
Now we will apply formula ${x_1},{x_2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 - 8} }}{2}$
Now we will simplify terms inside the square root
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt 1 }}{2}$
Here we can see that two roots will come so we will solve further to get roots of the equation. $ \Rightarrow x = \dfrac{{ - 3 \pm 1}}{2}$
$ \Rightarrow {x_1} = \dfrac{{ - 3 + 1}}{2} = \dfrac{{ - 2}}{2} = - 1$
$\therefore {x_1} = - 1$
$ \Rightarrow {x_2} = \dfrac{{ - 3 - 1}}{2} = \dfrac{{ - 4}}{2} = - 2$
$\therefore {x_2} = - 2$
Now we will find the roots of the quadratic equation ${x^2} + 4x + m = 0$ by applying the formula. Comparing this equation ${x^2} + 4x + m = 0$ with the general equation $a{x^2} + bx + c = 0$ Here a=1, b=4 and c=m \[x = \dfrac{{ - 4 \pm \sqrt {{4^2} - 4 \times 1 \times m} }}{{2 \times 1}}\]
Now we will solve this equation by simplifying further.
$ \Rightarrow x = \dfrac{{ - 4 \pm \sqrt {16 - 4m} }}{2}$
Now we will simplify terms inside the square root
$ \Rightarrow {x_1}' = \dfrac{{ - 4 + \sqrt {16 - 4m} }}{2}$ And ${x_2}' = \dfrac{{ - 4 - \sqrt {16 - 4m} }}{2}$
Now we will compare ${x_1} = - 1$ root of the first equation to the ${x_1}'$ root of another equation. $ \Rightarrow {x_1} = \dfrac{{ - 4 + \sqrt {16 - 4m} }}{2}$
$ \Rightarrow \dfrac{{ - 4 + \sqrt {16 - 4m} }}{2} = - 1$
Now we will cross multiply
$ \Rightarrow - 4 + \sqrt {16 - 4m} = - 2$
Now we will take constants to one side
\[ \Rightarrow \sqrt {16 - 4m} = - 2 + 4\]
\[ \Rightarrow \sqrt {16 - 4m} = 2\]
Now we will square both sides.
\[ \Rightarrow {(\sqrt {16 - 4m} )^2} = {(2)^2}\]
\[ \Rightarrow 16 - 4m = 4\]
\[ \Rightarrow 12 = 4m\]
\[ \Rightarrow m = \dfrac{{12}}{4} = 3\]
Now we will compare${x_2} = - 2$root of first equation with ${x_2}'$root of another equation. $ \Rightarrow {x_2} = \dfrac{{ - 4 - \sqrt {16 - 4m} }}{2}$
$ \Rightarrow \dfrac{{ - 4 - \sqrt {16 - 4m} }}{2} = - 2$
Now we will cross multiply
$ \Rightarrow - 4 - \sqrt {16 - 4m} = - 4$
\[ \Rightarrow - \sqrt {16 - 4m} = 0\]
Now we will square both sides.
\[ \Rightarrow {( - \sqrt {16 - 4m} )^2} = {(0)^2}\]
\[ \Rightarrow 16 - 4m = 0\]
\[ \Rightarrow 16 = 4m\]
\[ \Rightarrow m = \dfrac{{16}}{4} = 4\]
\[\therefore m = 4\]
Since we don’t know which root is equal to both the equations, we have to check both possibilities.

Therefore option (A) and (B) both options are correct.

Note: These types of questions are called multi optional questions. Some students only compare one root and forget about the rest root so this mistake should not be done as we don’t know which root of both equations is equal.