Problem 22 - Intermediate Algebra

Question

Solve. \(\sqrt{2 x-4}-\sqrt{3 x+4}=-2\)

Step-by-Step Solution

Answer

The solutions are \( x = 20 \) and \( x = 4 \).

Step 1: Isolate One Square Root

To simplify the equation, isolate one of the square roots on one side. We'll rearrange the equation: \( \sqrt{2x - 4} = \sqrt{3x + 4} - 2 \).

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square roots: \[ (\sqrt{2x - 4})^2 = (\sqrt{3x + 4} - 2)^2 \]. This results in: \( 2x - 4 = (3x + 4) - 4\sqrt{3x + 4} + 4 \).

Step 3: Simplify the Equation

Simplify the equation by combining like terms: \( 2x - 4 = 3x + 8 - 4\sqrt{3x + 4} \).

Step 4: Isolate the Square Root Again

Isolate the square root by moving all other terms to the other side: \( -x - 12 = -4\sqrt{3x + 4} \). Divide both sides by -4: \( \frac{x + 12}{4} = \sqrt{3x + 4} \).

Step 5: Square Both Sides Again

Square both sides again to fully eliminate the square root: \[ \left(\frac{x + 12}{4}\right)^2 = 3x + 4 \]. This results in: \( \frac{(x + 12)^2}{16} = 3x + 4 \).

Step 6: Clear the Fraction

Multiply through by 16 to remove the fraction: \( (x + 12)^2 = 48x + 64 \).

Step 7: Expand and Rearrange

Expand \((x + 12)^2\) using FOIL: \( x^2 + 24x + 144 \). Rearrange to form a quadratic equation: \( x^2 + 24x + 144 = 48x + 64 \).

Step 8: Solve the Quadratic Equation

Move all terms to one side of the equation to get: \( x^2 - 24x + 80 = 0 \). Solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -24 \), and \( c = 80 \). Calculate the discriminant: \( \sqrt{576 - 320} = \sqrt{256} = 16 \). The roots are \( x = \frac{24 \pm 16}{2} \).

Step 9: Find Values for x

Calculate the roots: \( x = \frac{40}{2} = 20 \) and \( x = \frac{8}{2} = 4 \).

Step 10: Verify Solutions

Verify by substituting back into the original equation. For \( x = 20 \): \( \sqrt{40 - 4} - \sqrt{60 + 4} = 6 - 8 = -2 \), true. For \( x = 4 \): \( \sqrt{8 - 4} - \sqrt{12 + 4} = 2 - 4 = -2 \), true.

Key Concepts

Isolating Square Roots

When faced with an equation involving square roots, the first step is to isolate one of the square root terms. This simplifies the equation and makes it easier to solve. For example, if we start with the equation \( \sqrt{2x - 4} - \sqrt{3x + 4} = -2 \), the goal is to have one square root by itself on one side of the equation. This can be achieved by rearranging terms, as such:

• Move one square root to the other side of the equation.
• For the equation, this looks like: \( \sqrt{2x - 4} = \sqrt{3x + 4} - 2 \).

By isolating the square root, we simplify our operations, making it easier to proceed with further algebraic manipulations. This is a crucial first step when solving these types of equations.

Squaring Both Sides

Squaring both sides of an equation is a common technique used to eliminate square roots. Once you've isolated a square root, the next step is to square each side. This "undoes" the square root and allows you to work with a polynomial instead. Consider the isolated equation \( \sqrt{2x - 4} = \sqrt{3x + 4} - 2 \):

• Square both sides to eliminate the square root: \( (\sqrt{2x - 4})^2 = (\sqrt{3x + 4} - 2)^2 \).
• This results in: \( 2x - 4 = 3x + 4 - 4\sqrt{3x + 4} + 4 \).

Squaring ensures that you are left with terms that can be simplified further, leading to the eventual solution of the equation. Note that after squaring both sides, further squaring might be necessary if not all roots are cleared immediately.

Quadratic Formula

Once the equation is freed of square roots and simplified, it often results in a quadratic equation. The quadratic formula is a powerful tool for finding the roots of any quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Here's how it's applied:

• Identify coefficients \( a \), \( b \), and \( c \).
• Plug these into the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In this problem, the quadratic equation is \( x^2 - 24x + 80 = 0 \), so:

• \( a = 1 \), \( b = -24 \), and \( c = 80 \).
• Find the discriminant, \( \sqrt{576 - 320} = 16 \).
• The solutions are \( x = \frac{24 \pm 16}{2} \), leading to \( x = 20 \) and \( x = 4 \).

The quadratic formula provides an efficient approach to determine potential solutions once the quadratic expression is set up.

Verifying Solutions

Verifying solutions is the final crucial step in making sure that the roots found satisfy the original equation. This step helps identify extraneous roots, which are solutions that might arise during the squaring process but do not actually solve the original equation.

• Substitute each solution back into the original equation to check validity.
• For \( x = 20 \), substitution gives: \( \sqrt{2 \times 20 - 4} - \sqrt{3 \times 20 + 4} = -2 \).
• Calculate: \( \sqrt{40 - 4} - \sqrt{60 + 4} = 6 - 8 = -2 \).
• For \( x = 4 \), substitution gives: \( \sqrt{2 \times 4 - 4} - \sqrt{3 \times 4 + 4} = -2 \).
• Calculate: \( \sqrt{8 - 4} - \sqrt{12 + 4} = 2 - 4 = -2 \).

Both solutions satisfy the original equation, verifying that they are correct. Always ensure to verify your solutions, especially when dealing with square roots, as extraneous roots can often occur.