Problem 27 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{3 t+7}-t=1 $$

Step-by-Step Solution

Answer

The solution is \(t = 3\); \(t = -2\) is extraneous.

Step 1: Isolate the square root

Start with the given equation: \[ \sqrt{3t+7} - t = 1 \] First, isolate the square root expression by adding \(t\) to both sides:\[ \sqrt{3t+7} = t + 1 \]

Step 2: Square both sides to eliminate the square root

Now that the square root is isolated, square both sides of the equation to eliminate the square root: \[ (\sqrt{3t+7})^2 = (t+1)^2 \]This leads to: \[ 3t + 7 = t^2 + 2t + 1 \]

Step 3: Rearrange into a quadratic equation

Rearrange the equation from the previous step into a standard quadratic form:\[ t^2 + 2t + 1 - 3t - 7 = 0 \] Simplifying gives:\[ t^2 - t - 6 = 0 \]

Step 4: Factor the quadratic equation

Now, we need to factor the quadratic equation:\[ t^2 - t - 6 \]Look for two numbers whose product is \(-6\) and whose sum is \(-1\). These numbers are \(-3\) and \(2\). Therefore, the equation factors to:\[ (t - 3)(t + 2) = 0 \]

Step 5: Solve for the roots of the equation

Set each factor equal to zero and solve for \(t\):\[t - 3 = 0 \Rightarrow t = 3 \]\[t + 2 = 0 \Rightarrow t = -2 \]

Step 6: Check the solutions for extraneous results

Substitute \(t = 3\) back into the original equation to check:\[ \sqrt{3(3) + 7} - 3 = 1 \]\[ \sqrt{16} - 3 = 1 \Rightarrow 4 - 3 = 1, \text{ which is true} \]Next, substitute \(t = -2\):\[ \sqrt{3(-2) + 7} - (-2) = 1 \]\[ \sqrt{1} + 2 = 1 \Rightarrow 1 + 2 = 1, \text{ which is false} \]Thus, \(t = -2\) is extraneous.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In the exercise, after isolating and squaring the equation \( \sqrt{3t+7} = t + 1 \), we derived a quadratic equation: \( t^2 - t - 6 = 0 \). Solving quadratic equations often involves rearranging terms to standard form, as was done in Step 3. Two common methods to solve these equations are factoring or using the quadratic formula. In our exercise, we chose the factoring method, looking for numbers that multiply to \(-6\) and add to \(-1\), thereby simplifying the process of solution finding.

Extraneous Solutions

Extraneous solutions are solutions that appear in the process of solving an equation but do not satisfy the original equation. These often arise during operations like squaring both sides of an equation, which can introduce additional roots. In our problem, after squaring both sides, we needed to check both potential solutions, \( t = 3 \) and \( t = -2 \). Only by substituting back into the original equation \( \sqrt{3t+7} - t = 1 \) can we confirm whether these solutions work.

• For \( t = 3 \), substitution gives a true statement: \( \sqrt{16} - 3 = 1 \).
• For \( t = -2 \), it becomes clear it does not satisfy the original equation: \( 1 + 2 eq 1 \).

This highlights the importance of validating solutions to avoid accepting extraneous solutions.

Factoring Methods

Factoring is a powerful technique used to break down algebraic expressions into simpler components, making them easier to solve. In quadratic equations, factoring converts the equation into a product of simpler linear factors. When an equation is set to zero, as in our exercise with \( t^2 - t - 6 = 0 \), we sought two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the middle coefficient).

• We identified these numbers as \(-3\) and \(2\), allowing us to factor the quadratic as \((t-3)(t+2) = 0\).
• Setting each factor equal to zero gives us the possible solutions: \( t = 3 \) and \( t = -2 \).

Knowing how to factor is essential for efficiently solving quadratic equations and discerning between true and extraneous solutions.