Question

$(x-3)^2 + |x-3| - 11 = 0$ the sum of solutions of equation is

• A. 5
• B. 6

Answer 1

Answer: (B)

6

Answer 2

The given equation is $(x-3)^2 + |x-3| - 11 = 0$. We can notice that $(x-3)^2 = |x-3|^2$. Let $y = |x-3|$. Since $|x-3|$ represents an absolute value, $y \ge 0$. Substituting $y$ into the equation, we get: $y^2 + y - 11 = 0$

This is a quadratic equation in $y$. We can solve for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=1$, $c=-11$. $y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-11)}}{2(1)}$ $y = \frac{-1 \pm \sqrt{1 + 44}}{2}$ $y = \frac{-1 \pm \sqrt{45}}{2}$ $y = \frac{-1 \pm 3\sqrt{5}}{2}$

We have two possible values for $y$: $y_1 = \frac{-1 + 3\sqrt{5}}{2}$ $y_2 = \frac{-1 - 3\sqrt{5}}{2}$

Since $y = |x-3|$, $y$ must be non-negative ($y \ge 0$). We only consider the valid solution for $y$:

$|x-3| = \frac{-1 + 3\sqrt{5}}{2}$

Since the right side $\frac{-1 + 3\sqrt{5}}{2}$ is positive, the equation $|x-3| = k$ where $k > 0$ has two solutions: $x-3 = k$ and $x-3 = -k$. Here, $k = \frac{-1 + 3\sqrt{5}}{2}$.

Case 1: $x-3 = \frac{-1 + 3\sqrt{5}}{2}$ $x_1 = 3 + \frac{-1 + 3\sqrt{5}}{2} = \frac{6}{2} + \frac{-1 + 3\sqrt{5}}{2} = \frac{6 - 1 + 3\sqrt{5}}{2} = \frac{5 + 3\sqrt{5}}{2}$

Case 2: $x-3 = -\left(\frac{-1 + 3\sqrt{5}}{2}\right)$ $x-3 = \frac{1 - 3\sqrt{5}}{2}$ $x_2 = 3 + \frac{1 - 3\sqrt{5}}{2} = \frac{6}{2} + \frac{1 - 3\sqrt{5}}{2} = \frac{6 + 1 - 3\sqrt{5}}{2} = \frac{7 - 3\sqrt{5}}{2}$

The solutions of the equation are $x_1 = \frac{5 + 3\sqrt{5}}{2}$ and $x_2 = \frac{7 - 3\sqrt{5}}{2}$.

The question asks for the sum of the solutions. Sum of solutions $= x_1 + x_2$ Sum $= \frac{5 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2}$ Sum $= \frac{(5 + 3\sqrt{5}) + (7 - 3\sqrt{5})}{2}$ Sum $= \frac{5 + 3\sqrt{5} + 7 - 3\sqrt{5}}{2}$ Sum $= \frac{12 + (3\sqrt{5} - 3\sqrt{5})}{2}$ Sum $= \frac{12 + 0}{2}$ Sum $= \frac{12}{2}$ Sum $= 6$

The sum of the solutions is 6.