Question

The sum of all the roots of \(|x-5|^{2}-|x-5|-6=0\) is (a) 10 (b) 6 (c) 0 (d) None

Short answer

The sum of all the roots of the given equation \(|x-5|^2 - |x-5| - 6 = 0\) is 11, so the answer is (d) None.

Step-by-step solution

Rewrite the equation with absolute value cases

Since |x-5| can represent two cases, (x-5) and -(x-5), the equation can be written as two separate cases. This allows us to avoid absolute values. Case 1: \(x-5 \ge 0\) Under this case, the equation becomes: \((x-5)^2 - (x-5) - 6 = 0\) Case 2: \(x - 5 < 0\) Under this case, the equation becomes: \((-(x-5))^2 - (-(x-5)) - 6 = 0\) Now we will solve for x for both cases.

Solve for the roots

In both cases, the equations are quadratic. Let's solve for x for both cases. Case 1: \((x-5)^2 - (x-5) - 6 = 0\) Expanding the equation, we get: \(x^2 - 10x + 25 - x + 5 - 6 = 0\) Simplifying, we have: \(x^2 - 11x + 24 = 0\) Factoring gives us: \((x-3)(x-8) = 0\) The roots in this case are 3 and 8. However, they are valid roots only if they satisfy the condition \(x-5 \ge 0\). 8 satisfies this condition, but 3 does not. So, for now, we have one root: 8. Case 2: \((-(x-5))^2 - (-(x-5)) - 6 = 0\) Expanding the equation, we get: \((x-5)^2 + (x-5) - 6 = 0\) This simplifies to: \(x^2 - 9x + 24 = 0\) Factoring gives us: \((x-3)(x-6) = 0\) The roots in this case are 3 and 6. However, they are valid roots only if they satisfy the condition \(x-5 < 0\). 3 satisfies this condition, but 6 does not. So, for now, we have one root in this case: 3.

Compute the sum of the roots

Now we have the roots 8 and 3. Adding them together, we get: \(8 + 3 = 11\) Therefore, the sum of all the roots of the given equation is 11, so the answer is (d) None.

Key concepts

Quadratic Equations

Quadratic equations are fundamental to algebra and are recognizable by their standard form, which is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. This form allows us to identify the highest power of the variable, which is 2, hence the name 'quadratic', derived from 'quad' meaning 'square'.

These equations can take various shapes when graphed, but they all form a parabola, which can either open upwards or downwards depending on the sign of \( a \). The roots or solutions of a quadratic equation are the values of \( x \) for which the equation equals zero. These roots can be real or complex and can represent key points on the graph of the equation, specifically where the parabola crosses the x-axis.

Finding Roots

There are several methods for finding the roots of a quadratic equation, including factoring, using the quadratic formula, or completing the square. The appropriate method often depends on the specific form of the quadratic equation presented. Factoring, for example, is often the easiest method when the equation can be neatly divided into two binomial expressions.

Factoring Quadratic Equations

Factoring is a valuable skill when solving quadratic equations, particularly when the equation is factorable into a product of two binomials. For instance, the quadratic equation \( ax^2 + bx + c = 0 \) can potentially be factored into \( (x - n)(x - m) = 0 \), where \( n \) and \( m \) are the roots of the equation.

To factor a quadratic equation, you look for two numbers that both add up to the coefficient \( b \) and multiply to the constant term \( c \). These numbers will form the 'inner' and 'outer' parts of the two binomials when expanded. Once factored, the equation is set to zero, and the Zero-Product Property is used, stating that if a product of factors equals zero, at least one of the factors must also be zero. This leads us directly to the roots of the equation.

However, factoring can be tricky, especially when the roots are not integers. In such cases, other methods like the quadratic formula may be necessary. Nonetheless, when feasible, factoring is often the quickest approach to finding the solutions to quadratic equations.

Solving Absolute Value Expressions

When it comes to absolute value expressions, it is important to understand that they measure the distance of a number from zero on the number line, regardless of direction. The absolute value of \( x \) is denoted by \( |x| \), and by definition, \( |x| \) is always nonnegative.

Solving equations with absolute value expressions involves considering two cases: one where the expression inside the absolute value is positive or zero, and another where it is negative. This stems from the piecewise definition of the absolute value function:

• If \( x \) is greater or equal to 0, then \( |x| = x \).
• If \( x \) is less than 0, then \( |x| = -x \).

This dual scenario leads to a split in the problem, where each case is handled separately as its own equation, often resulting in two different sets of solutions. After solving both cases, it is crucial to check which solutions are valid by plugging them back into the original equation to ensure they satisfy the absolute value conditions.

Solving absolute value equations can therefore sometimes lead to quadratic equations, as seen in the textbook problem, where the absolute value expressions were squared leading to quadratic forms that then had to be factored and solved accordingly.