Question

The sum of roots of the equation \[ |x - 1|^2 - 5 |x - 1| + 6 = 0 \] is

Hint:

For equations with absolute values, substitute the absolute value expression with a variable and solve the resulting equation.

Answer 1

Correct Answer: 4

To solve the given equation \(|x - 1|^2 - 5|x - 1| + 6 = 0\), we introduce a substitution: let \(y = |x - 1|\). The equation then becomes:

\[y^2 - 5y + 6 = 0\]

This is a quadratic equation in the standard form \(ax^2 + bx + c = 0\), where \(a=1\), \(b=-5\), and \(c=6\). We solve it using the quadratic formula:

\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Substituting the values, we get:

\[y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}\]

\[y = \frac{5 \pm \sqrt{25 - 24}}{2}\]

\[y = \frac{5 \pm \sqrt{1}}{2}\]

This gives us the roots:

\[y = \frac{5 + 1}{2} = 3\] and \[y = \frac{5 - 1}{2} = 2\]

Since \(y = |x - 1|\), we have two cases for each \(y\):

• For \(y = 3\):

\[|x - 1| = 3 \Rightarrow x - 1 = 3 \text{ or } x - 1 = -3\]

This gives \(x = 4\) or \(x = -2\).

• For \(y = 2\):

\[|x - 1| = 2 \Rightarrow x - 1 = 2 \text{ or } x - 1 = -2\]

This gives \(x = 3\) or \(x = -1\).

Thus, the roots of the original equation are \(x = 4, -2, 3, -1\).

The sum of the roots is:

\[4 + (-2) + 3 + (-1) = 4\]

Finally, the computed sum of roots \(4\) falls within the expected range [4, 4].