Problem 20 - A Graphical Approach to College Algebra

Question

Solve each equation by hand. Do not use a calculator. $$\sqrt[4]{2 x+3}=\sqrt{x+1}$$

Step-by-Step Solution

Answer

Solutions are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). Check for extraneous solutions.

Step 1: Eliminate Radicals

Our goal is to eliminate the radicals from the equation. To do this, raise both sides of the equation to the power of 4 (the index of the left side radical):\[ (\sqrt[4]{2x+3})^4 = (\sqrt{x+1})^4 \]Which simplifies to:\[ 2x + 3 = (x + 1)^2 \]

Step 2: Expand the Square

Now expand the right side of the equation:\[ 2x + 3 = (x+1)(x+1) \]This expansion results in: \[ 2x + 3 = x^2 + 2x + 1 \]

Step 3: Rearrange the Equation

Move all terms to one side of the equation to set it to zero:\[ 2x + 3 - 2x - 1 = x^2 + 2x + 1 - 2x - 1 \]This results in:\[ x^2 + 1 - 3 = 0 \]

Step 4: Simplify and Solve Quadratic Equation

Simplify the equation:\[ x^2 - 2 = 0 \]Solve for \(x\) by adding 2 to both sides:\[ x^2 = 2 \]Take the square root of both sides to find values for \(x\):\[ x = \pm \sqrt{2} \]

Step 5: Verify Solutions

Substitute the values back into the original equation to verify which, if any, are true solutions.Test \(x = \sqrt{2}\):- Left side: \( \sqrt[4]{2\sqrt{2}+3} \)- Right side: \( \sqrt{\sqrt{2}+1} \)- Check if both sides are equal.Test \(x = -\sqrt{2}\): Substitute the value and check.Verify that rational calculations confirm any potential extraneous solutions due to squaring.

Key Concepts

Understanding Radicals

Radicals are expressions that involve roots, such as square roots or cube roots. In this problem, we're dealing with a fourth root, which is a radical of order four.
To eliminate radicals, we raise both sides of the equation to the power corresponding to the radical. Here, since the left side contains a fourth root, we raise both sides of the equation to the fourth power.

• This step helps in simplifying the equation by getting rid of the radical sign.
• It transforms the radical equation into a polynomial equation, making it easier to solve.

The key is to apply the same power to both sides of the equation to maintain equality.
This technique is crucial when dealing with different types of root-based equations.

Solving Quadratic Equations

A quadratic equation is an equation where the highest exponent of the variable is two (e.g., \(x^2\)). In our problem, after eliminating the radicals, we are left with a quadratic equation:
\[ x^2 + 1 - 3 = 0 \]This simplifies to:\[ x^2 - 2 = 0 \]
To solve this, we first rearrange it in the standard quadratic form:

• Standard form: \( ax^2 + bx + c = 0 \)
• In this case, \(a = 1\), \(b = 0\), and \(c = -2\).

Once in this form, we can solve for \(x\) using the square root method, since our equation does not have a linear \(x\) term:\[ x = \pm \sqrt{2} \]Using the square root method is efficient when dealing with quadratics that do not need factoring or completing the square.

Verifying Solutions

Verifying solutions is an essential step after solving an equation, especially when dealing with radicals and exponents.
It's crucial because some operations, like squaring both sides of an equation, can introduce extraneous solutions—results that do not satisfy the original equation.
We need to check if our solutions, \(x = \sqrt{2}\) and \(x = -\sqrt{2}\), satisfy the original equation:\[ \sqrt[4]{2x+3} = \sqrt{x+1} \]To verify:

• Substitute \(x = \sqrt{2}\) into the original equation and simplify each side to ensure they're equal.
• Repeat the process for \(x = -\sqrt{2}\).

This verification step ensures any solutions we retain are true roots of the original equation, dismissing any that don’t hold to avoid errors.