Problem 62 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ 1=\sqrt{4 s+5}-\sqrt{2 s+2} $$

Step-by-Step Solution

Answer

The solutions are \(s = 1\) and \(s = -1\), and both are valid.

Step 1: Isolate One Square Root

To solve the equation \(1 = \sqrt{4s+5} - \sqrt{2s+2}\), first, isolate one of the square root terms. Let's move \(\sqrt{2s+2}\) to the other side to isolate \(\sqrt{4s+5}\):\[\sqrt{4s+5} = 1 + \sqrt{2s+2}.\]

Step 2: Eliminate the Square Root by Squaring

Square both sides of the equation to eliminate the square root on the left side:\[(\sqrt{4s+5})^2 = (1 + \sqrt{2s+2})^2.\]This simplifies to:\[4s+5 = 1 + 2\sqrt{2s+2} + 2s + 2.\]

Step 3: Simplify and Isolate the Remaining Square Root

Simplify the equation by combining like terms:\[4s + 5 = 2s + 3 + 2\sqrt{2s+2}.\]Subtract \(2s + 3\) from both sides to further isolate the square root:\[2s + 2 = 2\sqrt{2s+2}.\]

Step 4: Eliminate the Square Root Again by Squaring

Again, square both sides of the equation to remove the square root:\[(2s + 2)^2 = (2\sqrt{2s+2})^2.\]This simplifies to:\[4s^2 + 8s + 4 = 8s + 8.\]

Step 5: Solve the Quadratic Equation

Rearrange the equation to form a standard quadratic equation:\[4s^2 + 8s + 4 - 8s - 8 = 0,\]which simplifies to:\[4s^2 - 4 = 0.\]

Step 6: Factor and Solve for \(s\)

Factor the equation:\[4(s^2 - 1) = 0.\]This can be further factored as:\[4(s-1)(s+1) = 0.\]Therefore, the solutions are \(s = 1\) and \(s = -1\).

Step 7: Check for Extraneous Solutions

Substitute \(s = 1\) back into the original equation:\[1 = \sqrt{4(1)+5} - \sqrt{2(1)+2},\]\[1 = \sqrt{9} - \sqrt{4},\]\[1 = 3 - 2 = 1.\]This holds true.Now, substitute \(s = -1\):\[1 = \sqrt{4(-1)+5} - \sqrt{2(-1)+2},\]\[1 = \sqrt{1} - \sqrt{0},\]\[1 = 1 - 0 = 1.\]This holds true as well.

Key Concepts

square roots

Square roots are fundamental in mathematics and involve finding a number which, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, because 3 times 3 equals 9. In equations, square roots often appear with a symbol \(\sqrt{}\). To "undo" a square root in an equation, we typically square both sides. This is crucial for solving equations involving radicals.
In the given exercise, we start with an equation: \(1 = \sqrt{4s+5} - \sqrt{2s+2}\). It involves two square root terms. By manipulating the equation to isolate one of these terms, we can eliminate the square root by squaring both sides. This step is necessary to simplify and eventually solve for the variable.
Common tips for dealing with square roots in equations include:

• Isolate the square root term when possible to simplify manipulation.
• Remember that squaring both sides is reversible within certain constraints.
• Always check potential solutions in the original equation to verify, as squaring can introduce errors or extraneous solutions.

quadratic equations

Quadratic equations are polynomial equations of degree 2, typically written in the form \(ax^2 + bx + c = 0\). They are central in many areas of algebra and appear frequently in solving practical and theoretical problems.
In this exercise, by eliminating the square roots through squaring, a quadratic equation \(4s^2 - 4 = 0\) emerges. Solving a quadratic equation can be done using various methods like factoring, using the quadratic formula, or completing the square.
Key steps to solve the resultant quadratic equation include:

• Rearrange the equation to standard form.
• Factor, if possible: \(4(s - 1)(s + 1) = 0\).
• Solve for each factor equals zero: \(s = 1\) and \(s = -1\).

This understanding facilitates solving complex problems and lays the groundwork for understanding more advanced equations.

extraneous solutions

Extraneous solutions are potential solutions that arise during the algebraic manipulation of an equation but do not satisfy the original equation. They often occur in equations involving square roots or any form of manipulation that is not equivalent in all cases.
In this exercise, we initially obtain two solutions: \(s = 1\) and \(s = -1\). However, an important step in the final part of solving is verifying these solutions. This is done by substituting back into the original equation to ensure both solutions satisfy it. Upon substitution, both solutions actually prove valid as:

• Substitute \(s = 1\): \(1 = \sqrt{9} - \sqrt{4}\) \(\Rightarrow 1 = 3 - 2 = 1\).
• Substitute \(s = -1\): \(1 = \sqrt{1} - \sqrt{0}\) \(\Rightarrow 1 = 1 - 0 = 1\).

Understanding extraneous solutions is critical for ensuring accuracy and integrity when solving complex equations.