Problem 45 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(t+\sqrt{t}=12\)

Step-by-Step Solution

Answer

The solution is \( t = 9 \).

Step 1: Introduce a substitution

To simplify the equation, let’s introduce a substitution. Let’s set \( u = \sqrt{t} \). Then, \(t = u^2 \). Now the equation can be rewritten in terms of \(u\).

Step 2: Substitute and simplify

Using the substitution \(u = \sqrt{t}\), the original equation becomes \( u^2 + u = 12 \).

Step 3: Form and solve the quadratic equation

Rearrange the equation to form a standard quadratic equation:\( u^2 + u - 12 = 0 \). Solve this quadratic equation using the quadratic formula:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1\), \(b = 1\), and \(c = -12\).

Step 4: Apply the quadratic formula

Substitute the values into the quadratic formula:\[ u = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-12)}}{2(1)} \]Simplify inside the square root to get:\[ u = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm \sqrt{49}}{2} \]So we get two solutions:\[ u = \frac{-1 + 7}{2} = 3 \]\[ u = \frac{-1 - 7}{2} = -4 \]

Step 5: Revert the substitution

Recall that \( u = \sqrt{t} \). Hence, we have:For \( u = 3 \): \( 3 = \sqrt{t} \Rightarrow t = 9 \).For \( u = -4 \): \( -4 = \sqrt{t} \) is not valid as the square root of a real number cannot be negative.

Step 6: Verify the solution

Check the solution \( t = 9 \) in the original equation:\( t + \sqrt{t} = 9 + \sqrt{9} = 9 + 3 = 12 \), which is true. Therefore, \( t = 9 \) is the correct solution.

Key Concepts

Substitution Method

The substitution method in algebra involves replacing a variable with another expression to simplify an equation. In our case, we had the equation:
\( t + \sqrt{t} = 12 \)
This is challenging to solve directly because of the square root. To simplify, we introduced a new variable:
\( u = \sqrt{t} \).
Then, by squaring both sides, we get:
\( t = u^2 \).
By substituting \(u^2\) for \(t\) in the equation, it becomes easier to handle the algebraic expressions. This step transforms our original equation into a quadratic one:
\( u^2 + u = 12 \).

Quadratic Equations

A quadratic equation is any equation of the form:
\( ax^2 + bx + c = 0 \).
In our transformed equation, we have:
\( u^2 + u - 12 = 0 \).
Here, we recognize the standard quadratic form with:

• \(a = 1\)
• \(b = 1\)
• \(c = -12\)

Quadratic equations can have up to two solutions, found by various methods including factoring, completing the square, or using the quadratic formula. In this solution, we chose the quadratic formula as it directly provides the solutions even when factoring is not straightforward.

Quadratic Formula

The quadratic formula solves for \(x\) in a quadratic equation \( ax^2 + bx + c = 0 \) and is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \( u^2 + u - 12 = 0 \), substituting \(a = 1\), \(b = 1\), and \(c = -12\), we get:
\[ u = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-12)}}{2(1)} \]
Simplifying inside the square root:
\[ u = \frac{-1 \pm \sqrt{1 + 48}}{2} \]
\[ u = \frac{-1 \pm \sqrt{49}}{2} \]
Since \(\sqrt{49} = 7\), we have two solutions:
\[ u = \frac{-1 + 7}{2} = 3 \]
\[ u = \frac{-1 - 7}{2} = -4 \]
Thus, we get \(u = 3\) and \(u = -4\). However, since \(u = \sqrt{t}\), only positive solutions are valid, discarding \(u = -4\).

Solution Verification

After finding the possible solutions, it's crucial to verify them in the original equation to ensure they work. For our solution \( u = 3 \), reverting the substitution:
\( 3 = \sqrt{t} \), which means \( t = 9 \).
Substitute \( t = 9 \) back into the original equation to confirm:
\( 9 + \sqrt{9} = 9 + 3 = 12 \).
This satisfies the original equation, confirming that \( t = 9 \) is the correct solution. This final step of verification doesn't just provide confidence in our answer but illustrates the importance of returning to the initial equation to ensure all transformations hold true.