Problem 47 - Algebra A Combined Function

Question

Solve. $$ \sqrt{3 x+4}-1=\sqrt{2 x+1} $$

Step-by-Step Solution

Answer

Both solutions, \(x = 0\) and \(x = 4\), satisfy the original equation.

Step 1: Isolate the Square Roots

To start solving the equation \( \sqrt{3x + 4} - 1 = \sqrt{2x + 1} \), add 1 to both sides to isolate the square root on the left side: \[ \sqrt{3x + 4} = \sqrt{2x + 1} + 1 \]

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square roots. This gives:\[ (\sqrt{3x + 4})^2 = (\sqrt{2x + 1} + 1)^2 \]\[ 3x + 4 = (2x + 1) + 2\sqrt{2x + 1} + 1 \]Thus, we have:\[ 3x + 4 = 2x + 2 + 2\sqrt{2x + 1} \]

Step 3: Simplify and Isolate the Remaining Square Root

Rearrange the equation to isolate the remaining square root:\[ 3x + 4 - 2x - 2 = 2\sqrt{2x + 1} \]\[ x + 2 = 2\sqrt{2x + 1} \]

Step 4: Square Both Sides Again

Square both sides of the new equation:\[ (x + 2)^2 = (2\sqrt{2x + 1})^2 \]\[ x^2 + 4x + 4 = 4(2x + 1) \]\[ x^2 + 4x + 4 = 8x + 4 \]

Step 5: Solve the Quadratic Equation

Rearrange the equation to form a quadratic equation:\[ x^2 + 4x + 4 - 8x - 4 = 0 \]\[ x^2 - 4x = 0 \] Factor the quadratic:\[ x(x - 4) = 0 \]Thus, the solutions are \(x = 0\) and \(x = 4\).

Step 6: Verify the Solutions

Plug both solutions back into the original equation to check for validity. For \(x = 0\): \[ \sqrt{3(0) + 4} - 1 = \sqrt{2(0) + 1} \] \[ \sqrt{4} - 1 = \sqrt{1} \] \[ 2 - 1 = 1 \] This holds true.For \(x = 4\): \[ \sqrt{3(4) + 4} - 1 = \sqrt{2(4) + 1} \] \[ \sqrt{12 + 4} - 1 = \sqrt{8 + 1} \] \[ \sqrt{16} - 1 = \sqrt{9} \] \[ 4 - 1 = 3 \] This also holds true.

Key Concepts

Quadratic Equations

A quadratic equation is any equation of the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants and \(a\) is not equal to zero. These equations are called "quadratic" because they involve the square of the variable \(x\). Solving quadratic equations often involves finding the values of \(x\) that satisfy the equation. These solutions or "roots" can be found using various methods.

Some common methods include:

• Factoring
• Using the quadratic formula
• Completing the square

In our given problem, after manipulating the equation and eliminating the radicals, you isolate a quadratic equation \( x^2 - 4x = 0 \). This is a simplified form and can be easily solved by factoring, as we will see in the next section.

Factoring

Factoring is a method used to solve quadratic equations by expressing them as a product of simpler expressions, i.e., binomials. Once an equation is in its factored form, it's easy to set each factor equal to zero and solve for the variable \(x\). This method works best when the quadratic is straightforward and factorable.

For example, in \( x^2 - 4x = 0 \), you can factor by taking out the common factor \(x\). This leaves you with:

• \(x(x - 4) = 0\)

The solutions are found by setting each factor equal to zero:

• \(x = 0\)
• \(x - 4 = 0\), which means \(x = 4\)

Thus, there are two potential solutions: \(x = 0\) and \(x = 4\).

Verifying Solutions

After finding potential solutions for an equation, it's important to verify them. This means checking whether these solutions satisfy the original equation you started with. Verification is key in equations involving radicals, as squaring steps can sometimes introduce extraneous solutions that don't actually satisfy the original equation.

For our solutions \(x = 0\) and \(x = 4\):

• Start by substituting \(x = 0\) back into the original equation \( \sqrt{3x + 4} - 1 = \sqrt{2x + 1} \), which simplifies to \(2 - 1 = 1\). This holds true.
• Next, check \(x = 4\), which makes the original equation \(4 - 1 = 3\). This also holds true.

Both values satisfy the original equation, confirming them as valid solutions.