Find the two values of m ∈ ℤ for which the following equation in x has exactly one solution: $$3x^2 - mx + 3 = 0$$ Explain why the following equation in x has no real solutions: $$(2x + 3)^2 + 7 = 0$$ Show that x = -1 is not a solution of $$3x^2 + 2x + 5 = 0$$ - Leaving Cert Mathematics - Question 1 - 2022

The equation $(2x + 3)^2 + 7 = 0$ has no real solutions because the left side of the equation, $(2x + 3)^2$, is a perfect square and is always non-negative. Therefore:

$(2x + 3)^2 \\geq 0$ and adding 7 gives:

$(2x + 3)^2 + 7 \\geq 7 > 0$

Since the left side cannot equal zero, the equation has no real solutions.

To show that $x = -1$ is not a solution, we substitute $x = -1$ into the equation:

$3(-1)^2 + 2(-1) + 5 = 3(1) - 2 + 5 = 3 - 2 + 5 = 6$

Since the left-hand side equals 6, which is not equal to 0, it is confirmed that $x = -1$ is not a solution.

Using polynomial long division, we can express:

$3x^2 + 2x + 5 = (x + 1)(ax + b) + c$

To find the values of a, b, and c, we first evaluate the polynomial at $x = -1$:

$3(-1)^2 + 2(-1) + 5 = 3(1) - 2 + 5 = 6$

Thus, the remainder when $3x^2 + 2x + 5$ is divided by $x + 1$ is:

Remainder, c = 6.