Problem 43 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(3 x=\sqrt{16-10 x}\)

Step-by-Step Solution

Answer

\(x = \frac{8}{9}\)

Step 1: Isolate the square root term

First, isolate the square root term on one side of the equation: \[3x = \sqrt{16 - 10x}\]

Step 2: Square both sides

To eliminate the square root, square both sides of the equation: \[(3x)^2 = (\sqrt{16 - 10x})^2\] This simplifies to: \[9x^2 = 16 - 10x\]

Step 3: Rearrange into standard quadratic form

Rearrange the equation to form a standard quadratic equation:\[9x^2 + 10x - 16 = 0\]

Step 4: Factor or use the quadratic formula

Solve the quadratic equation using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a = 9\), \(b = 10\), and \(c = -16\). Plugging in these values:\[x = \frac{-10 \pm \sqrt{10^2 - 4(9)(-16)}}{2(9)} = \frac{-10 \pm \sqrt{100 + 576}}{18} = \frac{-10 \pm \sqrt{676}}{18}\] \[x = \frac{-10 \pm 26}{18}\]

Step 5: Solve for both potential solutions

This gives two potential solutions:\[x = \frac{16}{18} = \frac{8}{9}\] and \[x = \frac{-36}{18} = -2\]

Step 6: Check the solutions

Check both solutions in the original equation. For \(x = \frac{8}{9}\):\[3\left(\frac{8}{9}\right) = \sqrt{16 - 10\left(\frac{8}{9}\right)} \ \left(\frac{24}{9}\right) = \sqrt{16 - \left(\frac{80}{9}\right)} = \sqrt{\left(\frac{144}{9} - \frac{80}{9}\right)} = \sqrt{\frac{64}{9}} = \frac{8}{9}\] It hold true.For \(x = -2\):\[3(-2) = \sqrt{16 - 10(-2)} = \sqrt{16 + 20} = \sqrt{36} = 6 \ -6 eq 6\] Thus, \(x = -2\) is not a valid solution.

Key Concepts

Isolating Terms

When solving quadratic equations that include square root terms, the first step is to isolate the square root term. This means you need to get the square root by itself on one side of the equation.

By isolating the square root, you simplify the problem and make it easier to deal with subsequent steps.

Let's see how this applies to our example: \(3x=\sqrt{16-10x} \). Here, the square root term \(\sqrt{16 - 10x} \) is already isolated on one side of the equation. Isolating terms sets the stage for the next steps, like eliminating the square root by squaring both sides.

Using the Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a, b,\) and \(c\) are coefficients from the quadratic equation. In our example, after isolating and squaring, you get the equation \(9x^2 + 10x - 16 = 0\).

Plugging in these values: \(a = 9\), \(b = 10\), and \(c = -16\), the formula becomes:

\[ x = \frac{-10 \pm \sqrt{10^2 - 4(9)(-16)}}{2(9)} = \frac{-10 \pm \sqrt{676}}{18} = \frac{-10 \pm 26}{18} \]

This gives us two potential solutions: \[ x = \frac{16}{18} = \frac{8}{9} \] and \[ x = \frac{-36}{18} = -2 \]
These steps clearly show how to apply the quadratic formula to find solutions.

Checking Solutions

After finding the potential solutions, you must check them in the original equation to ensure they are valid.

For \( x = \frac{8}{9} \): \[ 3\left(\frac{8}{9}\right) = \sqrt{16 - 10\left(\frac{8}{9}\right)} \text{ which simplifies to } \left(\frac{24}{9}\right) = \sqrt{\left(\frac{144}{9} - \frac{80}{9}\right)} = \frac{8}{9} \]
\( x = \frac{8}{9} \) is valid.

For \( x = -2 \): \[ 3(-2) = \sqrt{16 - 10(-2)} = \sqrt{16 + 20} = \sqrt{36} = 6. \text{ Since -6 does not equal 6, } x = -2\, \text{ is not a valid solution.} \] This step eliminates any extraneous solutions and confirms the valid answers.