Problem 30 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{4-a}-a=2 $$

Step-by-Step Solution

Answer

The solution to the equation is \(a = 0\).

Step 1: Isolate the Square Root Term

Start by isolating the square root term on one side of the equation. The given equation is:\[ \sqrt{4-a} - a = 2 \]Add \(a\) to both sides:\[ \sqrt{4-a} = a + 2 \]

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square root. This gives:\[ (\sqrt{4-a})^2 = (a+2)^2 \]The left side becomes \(4-a\) and the right side needs to be expanded:\[ 4-a = a^2 + 4a + 4 \]

Step 3: Rearrange Terms

Bring all terms to one side of the equation to set it to zero:\[ a^2 + 4a + 4 - 4 + a = 0 \]Simplify the expression:\[ a^2 + 5a = 0 \]

Step 4: Factor the Quadratic

Factor the quadratic equation \(a^2 + 5a = 0\):\[ a(a + 5) = 0 \]Set each factor equal to zero to solve for \(a\):1. \(a = 0\)2. \(a + 5 = 0\) which gives \(a = -5\)

Step 5: Verify Solutions

Check if the proposed solutions \(a = 0\) and \(a = -5\) satisfy the original equation \( \sqrt{4-a} - a = 2 \):- For \(a = 0\):\[ \sqrt{4-0} - 0 = 2 \] simplifies to \(2 = 2\), which is true.- For \(a = -5\):\[ \sqrt{4-(-5)} - (-5) = 2 \] simplifies to \(3 + 5 = 2\), which is false.Thus, \(a = -5\) is extraneous and should be discarded.

Key Concepts

Extraneous Solutions

Sometimes when solving equations, especially those involving square roots, we encounter something called extraneous solutions. These are solutions that arise during the process of solving but do not actually satisfy the original equation. They're like impostors that fit all the steps until you try them in the starting line and realize something is off.

An extraneous solution often appears when we perform operations that might change the domain of the original equation, like squaring both sides. When you square a number, you lose information about the number's sign. This can create solutions that aren't valid for the initial equation.

• Always remember to check all potential solutions in the original equation.
• Discard any solutions that do not satisfy it.

In the given exercise, after solving, we found two solutions: \(a = 0\) and \(a = -5\). Substituting them back into the original equation, \(a = -5\) did not satisfy it, so it was extraneous. It's important to cross out such solutions to prevent errors in your work!

Squaring Both Sides

Squaring both sides of an equation is a common technique used to eliminate square roots. It involves raising both sides to the power of 2, thereby removing the square root. However, while it simplifies equations by getting rid of square roots, it comes with a catch.

This operation could introduce extraneous solutions, so caution is necessary. When you square both sides:

• Every positive and negative root becomes indistinguishable.
• Review the context of the equation to confirm the validity of your solutions.

In the exercise, the step of squaring both sides was used to turn \( \sqrt{4-a} = a + 2 \) into \(4-a = a^2 + 4a + 4\). While this got rid of the square root, it resulted in two solutions, from which only one turned out to be valid upon rechecking in the original equation.

Factoring Quadratics

Factoring quadratic equations is a valuable skill when solving for unknown variables. Quadratics come in the form \(ax^2 + bx + c = 0\), and factoring helps break it down into simpler parts. From there, solving becomes much more manageable.

The process involves:

• Finding factors of the quadratic that, when multiplied, return the original equation.
• Setting each factor equal to zero to solve for the variable.

In our exercise, after rearranging and simplifying, the quadratic \(a^2 + 5a = 0\) was factored into \(a(a + 5) = 0\). By setting each factor to zero, we found the potential solutions: \(a = 0\) and \(a = -5\). It’s a crucial step that helps simplify and solve the polynomial equations, leading us to solutions that we later verify.