Question

If \[a{x^2} + bx + c = 0\] and \[2{x^2} + 3x + 4 = 0\] have a common root, where \[a,b,c \in N\] (set of natural numbers), then the least value of \[a + b + c\] is:
A. \[13\]
B. $11$
C. $7$
D. $9$

Answer 1

Hint: As we know that \[a{x^2} + bx + c = 0\] and \[2{x^2} + 3x + 4 = 0\] have a common root so we have to first find the solution of \[2{x^2} + 3x + 4 = 0\] by applying the quadratic formula, then we calculate the value of \[a + b + c\]. We also use the theorem that if one of the roots of a quadratic equation is complex then the other root will be a conjugate of that root.

Complete step by step answer:
First we find the roots of \[2{x^2} + 3x + 4 = 0\] by applying quadratic formula that is \[x = \dfrac{{ - e \pm \sqrt D }}{{2d}}\] where $D$ is discriminant which is given by formula,\[D = {e^2} - 4df\] , and general equation is \[d{x^2} + ex + f = 0\] so on comparing given equation\[2{x^2} + 3x + 4 = 0\] and general equation, we get
\[d = 2\]
\[\Rightarrow e = 3\]
\[\Rightarrow f = 4\]

Substituting values of d, e and f in discriminant formula ,\[D = {e^2} - 4df\] , we get,
\[D = {\left( 3 \right)^2} - 4\left( 2 \right)\left( 4 \right)\]
\[ \Rightarrow D = 9 - 32\]
\[ \Rightarrow D = - 23\]
As the value of discriminant, D is negative so the roots of the given equation \[2{x^2} + 3x + 4 = 0\] will be complex. Now, substituting values of d, e and discriminant in quadratic formula \[x = \dfrac{{ - e \pm \sqrt D }}{{2d}}\], we get,
\[ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt { - 23} }}{{2\left( 2 \right)}}\]

Now, we can find both the roots by considering the positive and negative signs one by one.
So, considering the positive sign, we get,
\[x = \dfrac{{ - 3 + \sqrt { - 23} }}{4}\]
where \[\sqrt { - 1} = i\], so we get
\[ \Rightarrow x = \dfrac{{ - 3 + \sqrt {23} i}}{4}\]
Similarly, for negative sign, we get,
\[x = \dfrac{{ - 3 - \sqrt { - 23} }}{4}\]
where \[\sqrt { - 1} = i\], so we get
\[ \Rightarrow x = \dfrac{{ - 3 - \sqrt {23} i}}{4}\]
It is given that equation \[a{x^2} + bx + c = 0\] and \[2{x^2} + 3x + 4 = 0\] have a common root and roots of \[2{x^2} + 3x + 4 = 0\] are complex, which always exist in conjugate pair. So, both roots are common in both equations. This means the least value of \[a,b,c\] will be obtained at \[a = 2,b = 3,c = 4\]. So, the least value of \[a + b + c\] is \[2 + 3 + 4 = 9\].

Hence, option D is the correct answer.

Note: Given problem deals with solving the quadratic equation first to find its roots by using discriminant formula and then comparing the coefficients of both equations. The main thing to keep in mind while doing this question is to remember the theorem that complex roots always exist in conjugate pairs. Take care while doing the calculations.