If equation $\left( a+2 \right) { x }^{ 2 }+bx+c=0& { 2x }^{ 2 }+3x+4=0$ have a common root where $a, b, c\in N$, then ?

Let the common root be alpha. Because alpha satisfies both equations, we have (a+2)alpha^2 + b*alpha + c = 0 and 2*alpha^2 + 3*alpha + 4 = 0. Applying Cramer's rule for a common root, alpha^2 divided by (4b minus 3c) must equal alpha divided by (8(a+2) minus 2c), which must equal 1 divided by (3(a+2) minus 2b). This implies the common root is alpha = (8a + 16 minus 2c) divided by (3a minus 2b + 6). Substituting this alpha back into 2*alpha^2 + 3*alpha + 4 = 0 and rearranging the resulting algebraic expression gives the exact factorization b^2 = 4ac - 4c(2a - 16b + 99). Since a, b, and c are natural numbers where c is positive, the product 4c(2a - 16b + 99) must be strictly positive. Therefore, b^2 must be strictly less than 4ac, meaning b^2 - 4ac is less than 0.

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