Problem 422 - Elementary Algebra

Question

In the following exercises, solve. (a) \(\sqrt{u}+1=\sqrt{u+4}\) (b) \(\sqrt{n-5}+4=\sqrt{3 n+7}\)

Step-by-Step Solution

Answer

For part (a), there is no real solution. For part (b), the solutions are \(n = 14\) and \(n = 6\).

Step 1 Title - Isolate the square root

For part (a), start by isolating the square root on one side of the equation. Subtract 1 from both sides: \[ \sqrt{u} = \sqrt{u+4} - 1 \]

Step 2 - Square both sides

Square both sides of the equation to eliminate the square roots: \[ u = (\sqrt{u+4} - 1)^2 \]

Step 3 - Simplify the equation

Expand and simplify the right side of the equation: \[ u = (u+4) - 2\sqrt{u+4} + 1 \] Combine like terms:\[ u = u + 5 - 2\sqrt{u+4} \]

Step 4 - Isolate the square root again

Subtract \(u + 5\) from both sides: \[ 0 = -2\sqrt{u+4} \]

Step 5 - Solve for u

Divide by -2: \[ 0 = \sqrt{u+4} \]Square both sides again: \[ 0 = u+4 \]Finally, solve for \(u\):\[ u = -4 \]

Step 6 - Check for extraneous solutions

Verify by substituting back into the original equation. Substituting \(u = -4\) into \( \sqrt{u} + 1 = \sqrt{u+4} \) does not hold, making \(u = -4\) extraneous. Therefore, there is no real solution for part (a).

Step 7 - Repeat for part (b) – Isolate the square root

Start by isolating the square root on one side of the equation for part (b): \[ \sqrt{n - 5} + 4 = \sqrt{3n + 7} \]Subtract 4 from both sides: \[ \sqrt{n - 5} = \sqrt{3n + 7} - 4 \]

Step 8 - Square both sides

Square both sides of the equation to eliminate the square roots: \[ n - 5 = (\sqrt{3n + 7} - 4)^2 \]

Step 9 - Simplify the equation

Expand and simplify the right-hand side: \[ n - 5 = 3n + 7 - 8\sqrt{3n + 7} + 16 \] Combine like terms: \[ n - 5 = 3n + 23 - 8\sqrt{3n + 7} \]

Step 10 - Isolate the square root

Subtract \(3n + 23\) from both sides: \[ -2n - 28 = -8\sqrt{3n + 7} \]

Step 11 - Simplify further

Divide both sides by -8: \[ \sqrt{3n + 7} = \frac{2n + 28}{8} = \frac{n + 14}{4} \]

Step 12 - Square both sides again

Square both sides again to eliminate the remaining square roots: \[ 3n + 7 = \left(\frac{n + 14}{4}\right)^2 \]

Step 13 - Solve for n

Expand and simplify: \[ 3n + 7 = \frac{(n + 14)^2}{16} \]Multiply both sides by 16 to clear the fraction: \[ 48n + 112 = (n + 14)^2 \] Expand the right side: \[ 48n + 112 = n^2 + 28n + 196 \]Bring all terms to one side: \[ n^2 - 20n + 84 = 0 \] Use the quadratic formula to solve for n: \[ n = \frac{20 \pm \sqrt{400 - 336}}{2} = \frac{20 \pm \sqrt{64}}{2} = \frac{20 \pm 8}{2} \] The solutions are: \(n = 14\) and \(n = 6\)

Step 14 - Check for extraneous solutions

Verify by substituting back into the original equation. Substituting \(n = 14\) gives: \[ \sqrt{9} + 4 = \sqrt{49} \implies 3 + 4 = 7 \]True. Substituting \(n = 6\) gives: \[ \sqrt{1} + 4 = \sqrt{25} \implies 1 + 4 = 5 \]True. Thus, both solutions are correct.

Key Concepts

Square Root Isolation

When solving square root equations, the first step is to isolate the square root expression. This means you want to get the square root term by itself on one side of the equation.
For example, if you have an equation like \( \sqrt{u} + 1 = \sqrt{u + 4} \), you first need to subtract 1 from both sides:
\( \sqrt{u} = \sqrt{u + 4} - 1 \).
Isolating the square root makes the next steps easier because it sets you up to eliminate the square root by squaring both sides of the equation. Remember, the goal is to simplify your equation step-by-step until you can solve for the unknown variable.

Extraneous Solutions

Extraneous solutions are solutions that arise from the algebraic process but do not satisfy the original equation. When solving square root equations, squaring both sides can introduce these false solutions.
This is why it is important to always substitute your solutions back into the original equation to verify their validity.
For instance, in part (a) of our exercise, we found \( u = -4 \). Substituting this back into \( \sqrt{u} + 1 = \sqrt{u + 4} \) does not satisfy the equation, indicating \( u = -4 \) is extraneous.
Always check your solutions to ensure they are correct and not merely artifacts of the solving process.

Quadratic Formula

Quadratic equations often appear when solving square root problems. For example, when simplifying part (b), we end up with \( n^2 - 20n + 84 = 0 \).
To solve this, we use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a, b, \text{ and } c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
Plugging in our values from the equation, we get:
\[ n = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 84}}{2 \cdot 1} = \frac{20 \pm \sqrt{400 - 336}}{2} = \frac{20 \pm 8}{2} \]
This gives us the solutions \( n = 14 \) and \( n = 6 \), which upon checking both turned out to be correct.

Simplifying Equations

Simplifying equations involves breaking down complex expressions into simpler parts that are easier to work with.
In our example, after squaring both sides, we expanded and combined like terms:
For part (a), we had \( u = (u + 4) - 2 \sqrt{u + 4} + 1 \). Simplifying gave us \( u = u + 5 - 2 \sqrt{u + 4} \).
For part (b), simplifying \( \sqrt{n - 5} + 4 = \sqrt{3n + 7} \) led to more complex expressions, but by following step-by-step operations like distributing and combining like terms, we eventually isolated terms to solve for the variable. These steps turn large, complex equations into manageable pieces, making solving them straightforward.