Problem 27 - Algebra 2 and Trigonometry

Question

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ x+4 \sqrt{x}=5 $$

Step-by-Step Solution

Answer

The solution set is \(\{1\}\).

Step 1: Isolate the Square Root

To solve the equation \(x + 4\sqrt{x} = 5\), first try to isolate the square root term. We do this by subtracting \(x\) from both sides to get:\[ 4\sqrt{x} = 5 - x \]

Step 2: Solve for the Square Root

Now, divide both sides by 4 to solve for \(\sqrt{x}\):\[ \sqrt{x} = \frac{5 - x}{4} \]

Step 3: Square Both Sides

To eliminate the square root, square both sides of the equation:\[ x = \left(\frac{5 - x}{4}\right)^2 \]This becomes:\[ x = \frac{(5 - x)^2}{16} \]

Step 4: Clear the Denominator

Multiply both sides by 16 to clear the fraction:\[ 16x = (5 - x)^2 \]

Step 5: Expand the Quadratic

Expand \((5 - x)^2\):\[ 16x = 25 - 10x + x^2 \]

Step 6: Rearrange the Equation

Bring all the terms to one side of the equation to form a quadratic equation:\[ x^2 - 26x + 25 = 0 \]

Step 7: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -26\), and \(c = 25\):\[ x = \frac{-(-26) \pm \sqrt{(-26)^2 - 4 \cdot 1 \cdot 25}}{2 \cdot 1} \]\[ x = \frac{26 \pm \sqrt{676 - 100}}{2} \]\[ x = \frac{26 \pm \sqrt{576}}{2} \]\[ x = \frac{26 \pm 24}{2} \]Thus, the solutions are:\(x = \frac{50}{2} = 25\) and \(x = \frac{2}{2} = 1\).

Step 8: Check the Solutions

Substitute \(x = 25\) back into the original equation:\[ 25 + 4\sqrt{25} = 25 + 20 = 45 eq 5 \]This solution does not satisfy the original equation.Now, substitute \(x = 1\) into the original equation:\[ 1 + 4\sqrt{1} = 1 + 4 = 5 \]This solution satisfies the original equation, so \(x = 1\) is valid.

Step 9: Write the Solution Set

After checking, we find that the only valid solution is \(x = 1\). Therefore, the solution set is:\[ \{1\} \]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool that allows us to find the solutions to a quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here are the steps involved when using the quadratic formula:

• Identify the values of \(a\), \(b\), and \(c\) from the equation.
• Calculate the discriminant, \(b^2 - 4ac\).
• Substitute the values into the formula.
• Find the two possible values of \(x\) by using the plus-minus symbol \(\pm\).

Use the quadratic formula to solve the equation whenever it is difficult to factor or when precision is necessary, like in our solution where we found two potential solutions: \(x = 25\) and \(x = 1\).

Isolating Square Roots

Isolating the square root is about manipulating the equation to have the square root by itself on one side. This makes it easier to resolve the equation since getting rid of the square root requires squaring both sides. Here's a general approach:

• Move all other terms to the opposite side of the equation from the square root.
• Perform algebraic operations to leave the square root by itself.

In our original equation, \(x + 4\sqrt{x} = 5\), we isolated the square root by first subtracting \(x\) from both sides to get \(4\sqrt{x} = 5 - x\). Then, dividing both sides by 4, we achieved \(\sqrt{x} = \frac{5 - x}{4}\). This is an essential step before proceeding to eliminate the square root by squaring both sides.

Solution Set

Once you have found potential solutions to your quadratic equation, the next step is to determine which of these solutions are valid by constructing the solution set. The solution set consists of all values of the variable that satisfy the original equation. Here's how you determine it:

• After solving the equation, you may end up with multiple potential solutions.
• Substitute each solution back into the original equation to verify it actually works.

In our scenario, after using the quadratic formula on the equation \(x^2 - 26x + 25 = 0\), we calculated two solutions: 25 and 1. Upon checking, only \(x = 1\) satisfied the original equation, thus our solution set was \(\{1\}\).

Check Solutions

To check if the solutions you find from a quadratic equation are valid or not, you need to verify each solution by substituting it back into the original equation. This involves evaluating both sides of the equation with the given value of \(x\). Here's a step-by-step approach:

• Substitute each solution for \(x\) in the original equation.
• Calculate both sides of the equation to see if they are equal.

In our case, when we substituted \(x = 25\) back into \(x + 4\sqrt{x} = 5\), the equation did not hold true because 25 + 20 = 45, which is not equal to 5. Thus, 25 is not a valid solution. However, substituting \(x = 1\) resulted in both sides equaling 5, thus verifying it as a correct and valid solution for our solution set.