Solve 3x/2 - 9 = -x²/2: Multiplication-Free Solution Method

Let's solve the given equation:

$\frac{3x}{2}-9=-\frac{x^2}{2}$We will do this without multiplying both sides,

$$First, let's arrange the equation by moving terms:

$\frac{3x}{2}-9=-\frac{x^2}{2} \\ \frac{x^2}{2} +\frac{3x}{2}-9 =0 \\$Now, instead of multiplying both sides of the equation by the common denominator (to eliminate the fractions), we'll choose to factor out a fraction that's common to all terms. Note that the first two terms (which are squared and to the first power) are fractions, so we can use fraction multiplication and write the equation as follows:

$\frac{x^2}{2} +\frac{3x}{2}-9 =0 \\ \downarrow\\ \frac{1}{2}\cdot x^2+\frac{1}{2}\cdot 3x-9=0$

Next, we'll ask what number multiplied by $\frac{1}{2}$ gives us 9. The answer is obviously 18 since multiplication by one-half is equivalent to dividing by 2, and 18 is the result of multiplying 9 by 2:

$\textcolor{red}{18}\cdot\frac{1}{2}=\textcolor{red}{9\cdot2}\cdot\frac{1}{2}\\ 18\cdot\frac{1}{2}=9\cdot1\\ \downarrow\\ 18\cdot\frac{1}{2}=9$

Therefore, we can write the free number as:

$\frac{1}{2}\cdot x^2+\frac{1}{2}\cdot 3x-\textcolor{red}{9}=0\\ \downarrow\\ \frac{1}{2}\cdot x^2+\frac{1}{2}\cdot 3x-\textcolor{red}{\frac{1}{2}\cdot18}=0\\$

Let's continue, now we can factor out the common term $\frac{1}{2}$from all terms in the equation:

$\frac{1}{2}\cdot x^2+\frac{1}{2}\cdot 3x-\frac{1}{2}\cdot18=0\\ \downarrow\\ \frac{1}{2}\cdot(x^2+3x-18)=0\\$

Here we'll remember that the product of expressions equals zero only if at least one of the multiplying expressions equals zero,

However, the first factor in the expression we got is one-half, which is obviously not zero, therefore:

$x^2+3x-18=0$

Now we notice that in the resulting equation the coefficient of the quadratic term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-18\\ m+n=3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must be negative, therefore we can conclude that the numbers have different signs, according to multiplication rules. Now we'll remember that the possible factors of 18 are 2 and 9, 6 and 3, or 18 and 1. Meeting the second requirement mentioned, along with the fact that the numbers we're looking for have different signs leads to the conclusion that the only possibility for these two numbers is:

$\begin{cases} m=6\\ n=-3 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+3x-18=0 \\ \downarrow\\ (x+6)(x-3)=0$

Here we'll remember that the product of expressions equals zero only if at least one of the multiplying expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown in each:

$x+6=0\\ \boxed{x=-6}$

or:

$x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$\frac{3x}{2}-9=-\frac{x^2}{2} \\ \frac{x^2}{2} +\frac{3x}{2}-9 =0 \\ \downarrow\\ \frac{1}{2}\cdot(x^2+3x-18)=0\\ \downarrow\\ x^2+3x-18=0\\ \downarrow\\ (x+6)(x-3)=0 \\ \downarrow\\ x+6=0\rightarrow\boxed{x=-6}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=-6,3}$

Therefore the correct answer is answer D.