Solve the equation (1/3)(x - 3)^2 + (1/6)(x + 2)^2 = (x/2 - 1)(x/2 + 1) + (1/4)(x - 2)^2.

Step-by-step solution

1. 1

   Expand the left side. $(x-3)^2=x^2-6x+9$ and $(x+2)^2=x^2+4x+4$, so the left side is $\frac{1}{3}x^2-2x+3+\frac{1}{6}x^2+\frac{2}{3}x+\frac{2}{3}$.

2. 2

   Collect like terms on the left. $\frac13x^2+\frac16x^2=\frac12x^2$; $-2x+\frac23x=-\frac43x$; $3+\frac23=\frac{11}{3}$. The left side is $\frac12x^2-\frac43x+\frac{11}{3}$.

3. 3

   Expand the right side, using the difference of squares. $\left(\frac12x-1\right)\left(\frac12x+1\right)=\frac14x^2-1$, and $\frac14(x-2)^2=\frac14x^2-x+1$. Adding them gives $\frac12x^2-x$, since the constants $-1$ and $+1$ cancel.

4. 4

   Notice that the quadratic terms match. Both sides carry exactly $\frac12x^2$, so subtracting it leaves the linear equation $-\frac43x+\frac{11}{3}=-x$. This is why the problem is easier than its degree suggests: it is secretly linear.

5. 5

   Solve the linear equation. Add $\frac43x$ to both sides: $\frac{11}{3}=-x+\frac43x=\frac13x$. Multiplying by $3$ gives $x=11$.

6. 6

   Verify by substituting $x=11$. Left side: $\frac13(8)^2+\frac16(13)^2=\frac{64}{3}+\frac{169}{6}=\frac{297}{6}=\frac{99}{2}$. Right side: $\left(\frac{9}{2}\right)\left(\frac{13}{2}\right)+\frac14(9)^2=\frac{117}{4}+\frac{81}{4}=\frac{198}{4}=\frac{99}{2}$. The two sides agree exactly.