Problem 53 - Algebra 2

Question

Solve. Check for extraneous solutions. \(\sqrt{x+1}+\sqrt{2 x}=\sqrt{5 x+3}\)

Step-by-Step Solution

Answer

The solution for the equation is \(x=1\). The solution \(x=-1\) was found to be an extraneous solution when it was checked.

Step 1: Combine the roots on the left side

First, start by squaring the both sides to get rid of the radicals (square root signs) \((\sqrt{x+1}+\sqrt{2 x})^2=(\sqrt{5 x+3})^2\), this leads us to \(2\sqrt{(x+1)(2x)}+x+1+2x=5x+3\)

Step 2: Simplify

Next, simplify the equation by putting the terms in order, \(2\sqrt{(2x^2+2x)}=5x+3-x-1-2x = 2x+2\). Now we have \(2\sqrt{(2x^2+2x)}=2x+2\)

Step 3: Get Rid of the Remaining Root

Square both sides to get rid of the remaining root, giving \((2\sqrt{2x^2+2x})^2=(2x+2)^2\), this simplifies to \(8x^2 +8x=4x^2 +8x+4\)

Step 4: Simplify and Solve

Subtract \(4x^2+8x+4\) from both sides to set the equation to zero and solve quadratically, \(4x^2 = 4\). Divide by 4 to give \(x^2=1\). Rooting both sides now gives \(x=\pm1\).

Step 5: Check For Extraneous Solutions

Substitute \(x=1\) and \(x=-1\) back into the equation to verify. For \(x=1\), the equation holds true. For \(x=-1\), the equation is undefined, rendering it as an extraneous solution.

Key Concepts

Extraneous Solutions

When solving radical equations, it's crucial to check for extraneous solutions. These are solutions that appear legitimate mathematically but do not satisfy the original equation. This issue often arises when both sides of an equation are squared.

• Always substitute your solutions back into the original equation to verify their validity.
• If the values make the original equation undefined or false, they are extraneous.

If in our example, the solution for the equation r> oindent r> oindent r> oindent r> oindent becomes invalid, it suggests a misstep during the squaring process or the nature of radicals. Simplifying involves thoroughly checking if each solution works logically within the initial setup.

Squared Equations

Squaring an equation is a common strategy to eliminate radicals. While effective, it sometimes introduces extraneous solutions. This happens because squaring isn't a reversible operation.

• When you square both sides, you're introducing potentials for solutions that do not exist in the original equation.
• Square each side carefully and remember to follow through with recalculating to check if it matches the initial expression accurately.

For example, in the equation r> oindent r> oindent r> oindent r> oindent, each instance of squaring moves us further from the roots initially set. However, when used with caution, this tactic significantly simplifies radicals.

Simplifying Equations

Simplifying plays a vital role in solving equations efficiently. Simplification involves combining like terms, reducing complex fractions, and, importantly, removing radicals by suitable mathematical operations like squaring.

• Reorganize terms to make equations simpler and more approachable.
• Break down complex expressions into manageable segments.

In the example provided, simplifying the expression helps isolate the terms accurately which then can be squared again efficiently to achieve a simplified expression. Always cross-check if the rearranged terms maintain the equation's integrity.

Quadratic Solutions

After simplification and squaring, some equations become quadratic – taking the standard form of r> oindent r> oindent r> oindent. To solve these quadratic equations,

• Set the equation to equal zero.
• Factorize if possible, or use the quadratic formula—\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) .

In our exercise, solving the resulting quadratic equation involved factoring to find r> oindent r> oindent r> oindent . It's important to validate these solutions within the original problem context to avoid keeping extraneous solutions.