Problem 64 - Algebra for College Students

Question

Solve each equation by an appropriate method. $$y-1=\frac{\sqrt{y+1}}{2}$$

Step-by-Step Solution

Answer

\(y = \frac{9+\sqrt{33}}{8}\)

Step 1: Isolate the square root and square both sides

\(y - 1 = \frac{\sqrt{y+1}}{2}\). Multiply both sides by 2:
\(2(y-1) = \sqrt{y+1}\).
Square both sides: \(4(y-1)^2 = y+1\).

Step 2: Expand and solve the quadratic

\(4(y^2-2y+1) = y+1\)
\(4y^2-8y+4 = y+1\)
\(4y^2-9y+3 = 0\)

Using the quadratic formula: \(y = \frac{9 \pm \sqrt{81-48}}{8} = \frac{9 \pm \sqrt{33}}{8}\)

Step 3: Check for extraneous solutions

For \(y = \frac{9+\sqrt{33}}{8} \approx 1.843\): \(y-1 \approx 0.843 > 0\) and \(\frac{\sqrt{y+1}}{2} \approx \frac{\sqrt{2.843}}{2} \approx 0.843\). \(\checkmark\)

For \(y = \frac{9-\sqrt{33}}{8} \approx 0.407\): \(y-1 \approx -0.593 < 0\) but \(\frac{\sqrt{y+1}}{2} > 0\). \(\times\) (extraneous)

The solution is \(y = \frac{9+\sqrt{33}}{8}\).

Key Concepts

isolating the square root

Isolating the square root in an equation helps make solving it much simpler. Here's how you can do this: Start by identifying the term that contains the square root and try to get it by itself on one side of the equation. For example, in the equation \( y - 1 = \frac {\sqrt{y+1}}{2}\), we need to isolate the term \(\sqrt{y+1}\).

To do this, you need to first get rid of other terms on that side of the equation, such as constants or variables. In our case, add 1 to both sides of the equation to eliminate the constant term next to the variable y. This gives us:

\[ y - 1 + 1 = \frac {\sqrt{y+1}}{2} + 1 \]

Simplifying this, we have:

\[ y = \frac {\sqrt{y+1}}{2} + 1 \]

Notice that we still haven't completely isolated the square root. We need to get rid of the fraction. To do this, multiply both sides by 2:

\[ 2y = \sqrt{y+1} + 2 \]

Now we have successfully isolated the square root as the main term. This makes it easier for us to remove the radical in the next steps.

Always remember: isolating the square root first simplifies your equation and makes solving it more straightforward.

algebraic manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to isolate variables and solve equations. In our example, once we've isolated the square root, we're going to need to remove it.

Take the equation from the previous step:

\[ 2y = \sqrt{y+1} + 2\]

The next step involves getting rid of the square root by squaring both sides of the equation. Doing this helps us eliminate the radical term and makes solving the equation possible. Here’s how:

\[ (2y - 2)^2 = (\sqrt{y+1})^2 \]

This simplifies to:

\[ (2y - 2)^2 = y + 1 \]

At this point, you'll have an equation without any radicals that you can further simplify by expanding and combining like terms:

\[ 4y^2 - 8y + 4 = y + 1 \]

Subtract y and 1 from both sides to gather all terms to one side:

\[ 4y^2 - 8y + 4 - y - 1 = 0 \]

Combined, this looks like:

\[ 4y^2 - 9y + 3 = 0 \]

This new equation can be solved using factoring, the quadratic formula, or completing the square. The key point is that algebraic manipulation turns difficult terms (like square roots) into simpler forms that we can work with more easily.

radical equation

A radical equation is one in which the variable is found inside a radical, such as a square root. Solving these equations involves removing the radical to isolate and solve for the variable.

It's crucial to follow these steps:

• Isolate the radical on one side of the equation.
• Eliminate the radical by raising both sides of the equation to the power of the radical (usually squaring both sides if dealing with square roots).
• Simplify and solve the resulting equation.

In our specific example \(y - 1 = \frac{\sqrt{y + 1}}{2}\), after isolating and squaring, we got:

\[ 4y^2 - 9y + 3 = 0 \]

this resulted from eliminating the square root by squaring both sides.

Always be aware that squaring both sides can sometimes introduce extraneous solutions—values that solve the squared equation but not the original equation.

To ensure your solutions are valid, always plug them back into the original equation and see if they satisfy it. This way, you guarantee that the solutions are not just mathematically correct, but also contextually accurate for the problem presented.

By understanding and applying these steps to radical equations, you can systematically break them down and find correct solutions.