Problem 48 - Precalculus Student Solutions Manual 5th

Question

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

Step-by-Step Solution

Answer

The solution is \ \( x=2 \ \).

Step 1: Isolate one of the square roots

Begin by isolating one of the square root terms. Move \ \( \sqrt{x-1}\ \) to the right side of the equation: \ \( \sqrt{4x+1}=2+\sqrt{x-1}\ \)

Step 2: Square both sides

Next, square both sides of the equation to eliminate the square roots: \[ \left(\sqrt{4x+1}\right)^{2}=\left(2+\sqrt{x-1}\right)^{2}\ \]This simplifies to: \[ 4x+1=4+4\sqrt{x-1}+(x-1)\ \]Which further simplifies to: \[ 4x+1=3+x+4\sqrt{x-1}\ \]

Step 3: Isolate the remaining square root

Combine like terms on the right side to isolate \ \( \sqrt{x-1} \ \) again: \[ 4x+1=3+x+4\sqrt{x-1}\ \]Subtract \ \( x \) and \ \( 3 \) from both sides: \[ 3x-2=4\sqrt{x-1}\ \]

Step 4: Square both sides again

Square both sides again to fully eliminate the square root: \[ \left(3x-2\right)^{2}=\left(4\sqrt{x-1}\right)^{2}\ \]Simplifies to: \[ 9x^{2}-12x+4=16(x-1)\ \]Distribute and combine like terms: \[ 9x^{2}-12x+4=16x-16\ \]\[ 9x^{2}-28x+20=0\ \]

Step 5: Solve the quadratic equation

Now solve the quadratic equation \ \( 9x^{2}-28x+20=0 \ \) using the quadratic formula \ \( x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \): \ \( a=9, b=-28, c=20 \ \)\[ x = \frac{28\pm\sqrt{784-720}}{18}\ \]\[ x = \frac{28\pm8}{18}\ \]So, \[ x=2 \ \] or \ \( x = 10/9 \ \)

Step 6: Check for extraneous solutions

Substitute back to the original equation to check for extraneous solutions: For \ \( x=2 \ \): \[ \sqrt{4(2)+1}-\sqrt{2-1}=2\ \]\[ \sqrt{9}-1=2\ \]\[ 3-1=2\ \] which is true.For \ \( x=10/9 \ \): \[ \sqrt{4(10/9)+1}-\sqrt{10/9-1}=2\ \]Which will not be correct, thus \ \( x=10/9 \ \) is extraneous.

Key Concepts

Isolating Square Roots

To solve equations involving square roots, the first step is to isolate one of the square root terms on one side of the equation. This makes it easier to eliminate the square root by squaring both sides later. For example, in the equation \( \sqrt{4x + 1} - \sqrt{x - 1} = 2\), we start by moving \( \sqrt{x - 1}\) to the other side: \( \sqrt{4x + 1} = 2 + \sqrt{x - 1}\). Isolating the square root helps simplify the equation and prepares us for the next steps.

Squaring Both Sides

Once we have isolated one of the square roots, the next step is to square both sides of the equation to remove the square root. Squaring is effective because it turns the square root term into a simpler quadratic expression. In our example, squaring both sides of \( \sqrt{4x + 1} = 2 + \sqrt{x - 1}\) gives \( (\sqrt{4x + 1})^2 = (2 + \sqrt{x - 1})^2\). Simplifying this, we get \( 4x + 1 = 4 + 4\sqrt{x - 1} + x - 1\). Remember, when squaring both sides, always expand properly and combine like terms.

Solving Quadratic Equations

After squaring and simplifying, we typically end up with a quadratic equation. In our example, after further manipulation, we get \( 9x^2 - 28x + 20 = 0\). To solve a quadratic equation, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \( a\), \( b\), and \( c\) are coefficients from the quadratic equation \( ax^2 + bx + c = 0\). Substituting \( a = 9\), \( b = -28\), \( c = 20\), we solve for \( x\) to get the roots: \( x = 2\) or \( x = 10/9\).

Checking for Extraneous Solutions

Squaring both sides can sometimes introduce extraneous solutions. These are solutions that fit the squared equation but not the original one. Therefore, we always need to substitute the solutions back into the original equation to verify them. Substituting \( x = 2\) into the original equation \( \sqrt{4x + 1} - \sqrt{x - 1} = 2\), we find it to be true. However, for \( x = 10/9\), substituting back does not satisfy the original equation. Thus, \( x = 10/9\) is extraneous, and the only valid solution is \( x = 2\).