Question

If $x^2+3x+7=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c \in \mathbb{N}$ then minimum value of $(a+b+c)$

Answer 1

11

Answer 2

The discriminant of $x^2+3x+7=0$ is $D = 3^2 - 4(1)(7) = 9 - 28 = -19 < 0$. The roots are complex. Let $\gamma$ be the common root. Since $\gamma$ is a root of $x^2+3x+7=0$, we have $\gamma^2+3\gamma+7=0$, which implies $\gamma^2 = -3\gamma-7$. Since $\gamma$ is also a root of $ax^2+bx+c=0$, we have $a\gamma^2+b\gamma+c=0$. Substituting $\gamma^2$: $a(-3\gamma-7)+b\gamma+c=0$. Rearranging terms: $(b-3a)\gamma + (c-7a) = 0$. Since $\gamma$ is a complex number and $a,b,c$ are real (natural numbers), for this equation to hold, the coefficients of $\gamma$ and the constant term must both be zero. Thus, $b-3a=0 \implies b=3a$ and $c-7a=0 \implies c=7a$. We want to minimize $a+b+c = a+3a+7a = 11a$. Since $a,b,c \in \mathbb{N}$, the smallest possible value for $a$ is 1. When $a=1$, $b=3$, and $c=7$. These are all natural numbers. The minimum value of $(a+b+c)$ is $11 \times 1 = 11$.