Problem 441 - Elementary Algebra

Question

(a) Solve the equation \(\sqrt{r+4}-r+2=0\). (b) Explain why one of the "solutions" that was found was not actually a solution to the equation.

Step-by-Step Solution

Answer

The solution to the equation is \(\bbox[10px,border:2px solid #C0A000]{ 'r = 5'}\) while \(\bbox[10px,border:2px solid #C0A000]{`r = 0 }\) is an extraneous solution.

Step 1: Isolate the Square Root Term

Start by isolating the square root term on one side of the equation. Given the equation \(\bbox[10px,border:2px solid #C0A000]{\sqrt{r + 4} - r + 2 = 0}\), we can isolate the square root term by moving the other terms to the right-hand side: \(\bbox[10px,border:2px solid #C0A000]{\root[ ] r + 4 = r - 2}\)

Step 2: Square Both Sides

Square both sides to eliminate the square root. This yields: \(\bbox[10px,border:2px solid #C0A000]{ { (\root[ ] r + 4 ) = ( r - 2 )^2 } }\). Note that squaring both sides: \(\bbox[10px,border:2px solid #C0A000]{ r + 4 = ( r - 2 )^2 }\)

Step 3: Expand and Simplify

Expand and simplify the equation. \(\bbox[10px,border:2px solid #C0A000]{ r + 4 = ( r - 2 )( r - 2 ) = r^2 - 4r + 4 }\). This gives: \(\bbox[10px,border:2px solid #C0A000]{ r + 4 = r^2 - 4r + 4} }\).

Step 4: Move All Terms to One Side

Move all terms to one side of the equation so it equals zero: \(\bbox[10px,border:2px solid #C0A000]{ 0 = r^2 - 4r + 4 - r - 4 } }\). Simplify to: \(\bbox[10px,border:2px solid #C0A000]{0 = r^2 - 5r } }.\)

Step 5: Factor the Equation

Factor the equation: \(\bbox[10px,border:2px solid #C0A000]{ 0 = r ( r - 5 ) } \).

Step 6: Solve for r

Set each factor equal to zero: \(\bbox[10px,border:2px solid #C0A000]{ r = 0 } }\) and \(\bbox[10px,border:2px solid #C0A000]{ r - 5 = 0 } }.\)\ The solutions are \(\bbox[10px,border:2px solid #C0A000]{r = 0 \ , \ r = 5}\).

Step 7: Verify Solutions

Check each solution in the original equation. For \(\bbox[10px,border:2px solid #C0A000]{r = 0}\), \(\bbox[10px,border:2px solid #C0A000]{\root[ ] 4 - 0 + 2 eq 0 }\) as \(\bbox[10px,border:2px solid #C0A000]{\root[ ] 4 + 2 = 4 eq 0 }\). This is invalid. For \(\bbox[10px,border:2px solid #C0A000]{r=5}\), \(\bbox[10px,border:2px solid #C0A000]{ \root[ ] 9 - 5 + 2 eq 0 } \), it simplifies correctly as \(\bbox[10px,border:2px solid #3E9B00]{ \root[ ] 9 = 3 }\), which is correct. Therefore, valid solution is '5'.

Step 8: Explain Extraneous Solution

The extraneous solution occurred because squaring both sides of the equation can introduce solutions that are not valid in the original equation. \(\bbox[10px,border:2px solid #C0A000]{`r = 0 }\) was such a solution and did not satisfy the original equation.

Key Concepts

Isolating Square Roots

When solving square root equations, the first step is often to isolate the square root on one side of the equation. For example, if you start with the equation \(\bbox[{10px},{border:2px solid #C0A000}]{\sqrt{r + 4} - r + 2 = 0}\), you'll need to move all other terms to the opposite side to get \(\bbox[{10px},{border:2px solid #C0A000}]{\sqrt{r + 4} = r - 2}\). This simplification is crucial because it sets the stage for the next step: eliminating the square root by squaring both sides.

Squaring Both Sides

Once the square root term is isolated, the next step involves squaring both sides of the equation. This is done to eliminate the square root. For instance, starting from \(\bbox[{10px},{border:2px solid #C0A000}]{\sqrt{r + 4} = r - 2}\), squaring both sides results in \(\bbox[{10px},{border:2px solid #C0A000}]{r + 4 = (r - 2)^2}\). The square root is eliminated, but remember that squaring an equation can sometimes introduce extraneous solutions, which we'll discuss further.

Extraneous Solutions

Extraneous solutions are false solutions that arise when both sides of an equation are squared. In our example, after squaring both sides and solving, we found two potential solutions, \(\bbox[{10px},{border:2px solid #C0A000}]{r = 0\)\text{ and}\(r = 5}\). However, when we checked these in the original equation, we found that \(\bbox[{10px},{border:2px solid #C0A000}]{r = 0}\) did not satisfy it. This is a classic case of an extraneous solution: it appears valid in the squared equation but does not work in the original.

Algebraic Manipulation

Algebraic manipulation involves using various algebraic techniques to transform an equation into a simpler form. In this example, after isolating the square root and squaring both sides, we expanded and simplified to find \(\bbox[{10px},{border:2px solid #C0A000}]{r + 4 = r^2 - 4r + 4}\). Further simplification led to moving all terms to one side: \(\bbox[{10px},{border:2px solid #C0A000}]{0 = r^2 - 5r}\). Finally, we factored this to get \(\bbox[{10px},{border:2px solid #C0A000}]{0 = r(r - 5)}\) and found the solutions. Each of these steps involves careful algebraic manipulation to maintain the integrity of the solution.