Problem 41 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(2 x=\sqrt{11 x+3}\)

Step-by-Step Solution

Answer

The solution is \(x=3\).

Step 1 - Isolate the square root

First, isolate the square root on one side of the equation. The equation is already in the form where the square root is isolated, so it retains the form: \(2x = \sqrt{11x + 3}\)

Step 2 - Square both sides

To eliminate the square root, square both sides of the equation: \[(2x)^2 = (\sqrt{11x + 3})^2\] This simplifies to: \[4x^2 = 11x + 3\]

Step 3 - Rearrange into a standard quadratic equation

Rearrange the equation into the standard quadratic form \(ax^2 + bx + c = 0\): \[4x^2 - 11x - 3 = 0\]

Step 4 - Solve the quadratic equation using the quadratic formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=4\), \(b=-11\), and \(c=-3\). Calculate the discriminant first: \[b^2 - 4ac = (-11)^2 - 4(4)(-3) = 121 + 48 = 169\]Then, apply the quadratic formula: \[x = \frac{11 \pm \sqrt{169}}{8}\] This results in: \[x = \frac{11 + 13}{8} = \frac{24}{8} = 3\] \[x = \frac{11 - 13}{8} = \frac{-2}{8} = -\frac{1}{4}\]

Step 5 - Check the solutions

Substitute the solutions back into the original equation to check for extraneous solutions. For \(x = 3\): \[2(3) = \sqrt{11(3) + 3}\]\[6 = \sqrt{33 + 3}\]\[6 = \sqrt{36}\]\[6 = 6\] This is correct. For \(x = -\frac{1}{4}\): \[2 \left( -\frac{1}{4} \right) = \sqrt{11 \left( -\frac{1}{4} \right) + 3}\]\[ -\frac{1}{2} = \sqrt{ -\frac{11}{4} + 3}\]\[ -\frac{1}{2} = \sqrt{ -\frac{11}{4} + \frac{12}{4}}\]\[ -\frac{1}{2} = \sqrt{ \frac{1}{4} }\]\[ -\frac{1}{2} = \frac{1}{2}\] This is incorrect since it contradicts the equation.

Key Concepts

Square Root Isolation

Starting with the equation given: \[2x = \sqrt{11x + 3}\] We first perform an essential step called square root isolation. This step involves ensuring the square root is by itself on one side of the equation. In this case, the equation already has the square root isolated, so we can move on to squaring both sides.

Quadratic Formula

Once we have a standard quadratic equation in the form \[ax^2 + bx + c = 0\] we can solve it using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In our specific problem, after moving all terms to one side, we have: \[4x^2 - 11x - 3 = 0\]. Here, \[a = 4\], \[b = -11\], \[c = -3\]. Plugging these values into the quadratic formula helps us get the solutions.

Extraneous Solutions

Extraneous solutions are solutions that arise from manipulating the equation, especially when squaring both sides. These solutions do not satisfy the original equation. After solving the quadratic equation, we obtain potential solutions \[x = 3\] and \[x = -\frac{1}{4}\]. It is crucial to substitute these solutions back into the original equation to determine if they work. In our case, substituting \[x = 3\] returns a valid solution, but \[x = -\frac{1}{4}\] does not because it contradicts the original equation. This makes \[x = -\frac{1}{4}\] an extraneous solution.

Discriminant

The discriminant in the quadratic formula is the part under the square root, defined as: \[b^2 - 4ac\]. This value helps determine the nature of the roots of the quadratic equation:

• If \[b^2 - 4ac > 0\], there are two distinct real roots.
• If \[b^2 - 4ac = 0\], there is exactly one real root (a repeated root).
• If \[b^2 - 4ac < 0\], there are no real roots (the solutions are complex).

In our example, we calculated: \[ (-11)^2 - 4(4)(-3) = 121 + 48 = 169\]. Since 169 is greater than 0, we confirm there are two distinct real roots for our equation.