Problem 35 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ 2 x=\sqrt{11 x+3} $$

Step-by-Step Solution

Answer

The solution is \(x=3\). \(x=-\frac{1}{4}\) is extraneous.

Step 1: Eliminate the Square Root

To eliminate the square root, square both sides of the equation. This will allow us to work with a polynomial equation:\[(2x)^2 = (\sqrt{11x + 3})^2\]This simplifies to:\[4x^2 = 11x + 3\]

Step 2: Rearrange the Equation

Move all terms to one side of the equation to set it to zero. This will help you factor or apply the quadratic formula:\[4x^2 - 11x - 3 = 0\]

Step 3: Factor or Use the Quadratic Formula

Check if you can factor the quadratic. In this case, factoring is challenging, so use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our equation, let\(a = 4\),\(b = -11\), and\(c = -3\). Substitute these into the formula:\[x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}\]

Step 4: Simplify the Expression

Compute each part of the quadratic formula:\[b^2 = (-11)^2 = 121,\]\[4ac = 4 \cdot 4 \cdot (-3) = -48\]\[\sqrt{b^2 - 4ac} = \sqrt{121 + 48} = \sqrt{169} = 13\]Substitute back:\[x = \frac{11 \pm 13}{8}\]Calculate both solutions for \(x\):\[x_1 = \frac{11 + 13}{8} = 3,\]\[x_2 = \frac{11 - 13}{8} = -\frac{1}{4}\]

Step 5: Check for Extraneous Solutions

Substitute both values back into the original equation to ensure they satisfy it.For \(x = 3\):\[2(3) = \sqrt{11(3) + 3} \]\[6 = \sqrt{36} \]This is true. For \(x = -\frac{1}{4}\):\[2(-\frac{1}{4}) = \sqrt{11(-\frac{1}{4}) + 3}\]\[-\frac{1}{2} = \sqrt{\frac{-11}{4} + \frac{12}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\]This does not hold true, making \(x = -\frac{1}{4}\) an extraneous solution.

Key Concepts

Square Root Elimination

When dealing with equations that have square roots, the aim is often to eliminate the square root to simplify the problem. To do this, the most direct method is to square both sides of the equation. This technique removes the square root by taking advantage of the squared operation cancelling out the root.

In our example, we start with the equation: \[2x = \sqrt{11x + 3}\]Squaring both sides gives us:\[(2x)^2 = (\sqrt{11x + 3})^2\]Leading to:\[4x^2 = 11x + 3\]
This operation allows us to turn the original equation into a polynomial one, which is generally easier to solve using algebraic methods. However, it's essential to be aware that squaring can introduce extraneous solutions, which need later verification.

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). If factoring is cumbersome or impractical, this formula provides a way to find solutions systematically.

The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the roots (solutions) for any quadratic equation by substituting the specific values of \(a\), \(b\), and \(c\).
In our example:\[a = 4, \ b = -11, \ c = -3\]Plug these into the quadratic formula:\[x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}\]
After simplifying, we get:\[x_1 = 3, \ x_2 = -\frac{1}{4}\]This process ensures every potential solution is considered, even when the quadratic does not factor neatly.

Extraneous Solutions

Extraneous solutions are solutions that emerge from the process of solving the equation but do not satisfy the original equation.

This commonly occurs when squaring both sides of an equation, as additional solutions might be introduced that aren't valid for the original equation.
To identify these, it's crucial to substitute all solutions back into the original equation.
In our example:

• For \(x = 3\), substituting back shows the equality holds true.
• For \(x = -\frac{1}{4}\), substituting back results in a mismatch, indicating it's an extraneous solution.

The correct approach requires verifying each solution to ensure conformity with the original equation conditions.

Polynomial Equations

Polynomial equations are equations involving powers of variables. They generally follow the form \(ax^n + bx^{n-1} + \ldots + k = 0\), where \(n\) is a non-negative integer.

In our exercise, once the square root is eliminated through squaring, we end up with:\[4x^2 - 11x - 3 = 0\]This is a typical quadratic polynomial equation, which can be approached by various algebraic techniques like factoring or using the quadratic formula.
Polynomial equations are fundamental in algebra, and understanding them is necessary to tackle a variety of mathematical problems.
They allow us to describe relationships and changes, laying the groundwork for broader applications in math and science.