Problem 63 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solving an Equation Involving an Absolute Value Find all solutions of the equation algebraically. Check your solutions. $$|x+1|=x^{2}-5$$

Step-by-Step Solution

Answer

The solution to the absolute value equation \(|x+1|=x^{2}-5\) are \(x=3\) and \(x=-3\).

Step 1: Express Absolute Value Equation as Two Separate Equations

The principal of the absolute value is that if \(|a| = b\), then \(a\) can be \(b\) or \(-b\). In our case, from \(|x+1|=x^{2}-5\), we get two equations: \(x+1 = x^{2}-5\) or \(x+1 = -(x^{2}-5)\).

Step 2: Simplify The Two Equations

Reshape each equation into standard quadratic form. The first equation becomes \(x^{2}-x-6=0\), and the second equation becomes \(x^{2}+x-6=0\).

Step 3: Solve The Quadratic Equations

We now have 2 quadratic equations to solve. We can use the quadratic formula \(x={-b±\sqrt{b^{2}-4ac}}/{2a}\). For the first equation, \(a=1, b=-1, c=-6\), for the second equation, \(a=1, b=1, c=-6\). By substituting the values of a, b and c into the quadratic formula, we get the roots of the equations. The solutions for the first equation are \(x=3, -2\) and for the second equation, \(x=-3, 2\).

Step 4: Check The Solutions

Substitute the possible solutions we got from quadratic equations back into the original absolute equation to confirm. In this case, for the equation \(|x+1|=x^{2}-5, x=3\) and \(x=-3\) are the valid solutions. So, the solution of the given equation are \(x=3\) and \(x=-3\).

Key Concepts

Understanding Absolute Value Equations

Absolute value equations can initially seem a bit tricky, but they become clearer once you understand the basic concept. The absolute value of a number is its distance from zero, regardless of direction on a number line. So, for any number or expression inside absolute value brackets, such as \(|x+1|\), the result is always a positive value or zero. To solve absolute value equations, we split them into two possible cases:

• One where the expression inside equals the positive value of the other side.
• Another where it equals the negative value.

This approach helps in addressing both scenarios: when the inside value is positive, and when it is negative. Keep this in mind as you solve similar equations, and remember to handle each case separately.

Solving Quadratic Equations

Once we have split the absolute value equation into two cases, we often end up with quadratic equations. Quadratic equations usually take the form of \(ax^2 + bx + c = 0\). These can be solved using various methods including factoring, completing the square, or using the quadratic formula. Let's focus on the quadratic formula, which is \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]This formula is a reliable method to find the solutions or roots of any quadratic equation, as long as you substitute the correct values of \(a\), \(b\), and \(c\). In our example, solving \(x^2 - x - 6 = 0\) and \(x^2 + x - 6 = 0\) by plugging values into the formula yields roots that are potential solutions of the original problem.

Finding Algebraic Solutions

Algebraic solutions involve manipulating algebraic expressions to find values of unknown variables. To find solutions algebraically, follow these general steps:

• Isolate the variable by performing inverse operations.
• For equations involving absolute values, create separate cases based on possible scenarios of the expression inside the absolute value.
• Apply appropriate algebraic methods to simplify and solve each case.

It is essential to keep your work organized, as multiple possible solutions might arise, especially in quadratic equations. These solutions provide critical insight into understanding the behavior of algebraic expressions and how they equate mathematically.

Checking Solutions for Accuracy

Once potential solutions are determined algebraically, it's vital to verify them by substituting back into the original equation. This step confirms whether the solutions satisfy the equation fully. In our example, substitute the values \(x = 3\) and \(x = -3\) back into the original equation \(|x+1| = x^2 - 5\) to see if both sides equal. Checking solutions ensures:

• The integrity of your solution process is maintained.
• Any extraneous solutions, which don't actually satisfy the initial equation, are eliminated.

This step is crucial in validating your findings and cementing confidence in your algebraic skills.