Problem 57 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{z+3}-\sqrt{z}=1 $$

Step-by-Step Solution

Answer

The solution to the equation is \( z = 1 \).

Step 1 - Isolate One Square Root

To begin solving the equation \( \sqrt{z+3} - \sqrt{z} = 1 \), isolate one of the square roots. Let's add \( \sqrt{z} \) to both sides of the equation:\[ \sqrt{z+3} = 1 + \sqrt{z} \].

Step 2 - Square Both Sides to Eliminate the Square Root

Square both sides of the equation to eliminate the square root:\( (\sqrt{z+3})^2 = (1 + \sqrt{z})^2 \).This simplifies to:\[ z + 3 = 1 + 2\sqrt{z} + z \].

Step 3 - Simplify the Equation

Subtract \( z \) from both sides to simplify:\( 3 = 1 + 2\sqrt{z} \).Subtract 1 from both sides:\[ 2 = 2\sqrt{z} \].

Step 4 - Solve for the Square Root

Divide both sides by 2 to solve for \( \sqrt{z} \):\( \sqrt{z} = 1 \).

Step 5 - Solve for z

Now square both sides again to find \( z \):\( (\sqrt{z})^2 = 1^2 \). This leads to:\[ z = 1 \].

Step 6 - Verify the Solution

Substitute \( z = 1 \) back into the original equation to verify it is a valid solution:\( \sqrt{1+3} - \sqrt{1} = 1 \).Simplifying gives \( \sqrt{4} - 1 = 1 \), which simplifies to \( 2 - 1 = 1 \).Since this is true, \( z = 1 \) is a valid solution.

Key Concepts

Square Root

A square root is a value that, when multiplied by itself, gives the original number. For example, the square root of 4 is 2, because 2 multiplied by 2 equals 4. In equations, like the one in the exercise, square roots are often used to simplify expressions and to find solutions for variables.

When dealing with square roots in equations, a common goal is to eliminate them. This typically involves isolating the square root on one side of the equation, then squaring both sides. Squaring both sides turns the square root into a simpler term, facilitating further solution steps. However, one must be careful when squaring both sides to ensure that no information is lost and that no extraneous solutions are introduced.

Extraneous Solutions

Extraneous solutions may arise during the process of solving equations, especially those involving square roots or other radical expressions. These are solutions that fit the transformed equation, but not the original equation. Understanding their emergence is crucial.

Extraneous solutions typically appear when you square both sides of an equation. Squaring is not a reversible operation, meaning that not all transformations are equivalent in both directions. This can result in solutions that satisfy the squared equation, but not the original one. That's why it's important to check each solution by substituting it back into the original equation. If both sides equal, the solution is valid; otherwise, it's extraneous.

Equation Solving

Equation solving is all about finding the values of the variables that satisfy the given equation. It involves a series of logical steps that simplify the equation, making the solution clearer.

For the equation involving square roots, the process often starts with isolation of one root, then transforming the equation by squaring both sides. This process eliminates square roots and creates a polynomial equation, which is easier to manage.

The solution steps involve simplifying this new equation by collecting like terms and solving for the variable. It's essential to verify each prospective solution by plugging it back into the original equation to ensure it is valid, avoiding concluding with an extraneous solution.

• Isolate the variable-related terms.
• Eliminate square roots by squaring.
• Simplify and solve for the variable.
• Verify solutions against the original equation.

By methodically following these steps, you can effectively tackle and solve many types of equations, including those with radicals.