Problem 391 - Elementary Algebra

Question

In the following exercises, check whether the given values are solutions. For the equation \(\sqrt{t+6}=t\) (a) Is \(t=-2\) a solution? (b) Is \(t=3\) a solution?

Step-by-Step Solution

Answer

(a) No, (b) Yes

Step 1: Verify if t = -2 is a solution

Substitute \( t = -2 \) into the equation \( \sqrt{t+6}=t \). This gives \( \sqrt{-2+6}=-2 \). Simplifying inside the square root, \( \sqrt{4}=-2 \). Since \( \sqrt{4}=2 \), \( 2 e -2 \). Therefore, \( t = -2 \) is not a solution.

Step 2: Verify if t = 3 is a solution

Substitute \( t = 3 \) into the equation \( \sqrt{t+6}=t \). This gives \( \sqrt{3+6}=3 \). Simplifying inside the square root, \( \sqrt{9}=3 \). Since \( \sqrt{9}=3 \), which means \( 3=3 \). Therefore, \( t = 3 \) is a solution.

Key Concepts

Square Roots

Understanding square roots is crucial in solving equations like the one given in the exercise. A square root of a number 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. Notice that the square root of a number is always non-negative. In mathematical notation, we write the square root of a number as \( \sqrt{a} \). So, \( \sqrt{4} = 2 \).

• Any number inside the square root symbol must be non-negative.
• When simplifying a square root, ensure the inside value (called the radicand) is simplified first.
• The result of a square root operation is always positive, which we call the principal square root.

In our exercise, we had to check the value of \( t \) for which \( \sqrt{t+6} = t \). Let's see how verifying solutions is connected to this.

Verifying Solutions

Verifying solutions involves substituting the values back into the original equation to check if they work. It's like double-checking our math. Here's how we do it:

Step 1: Substitute the given value into the equation. This means if we have the value of \( t \), we put it in every place where we see \( t \) in the equation.

Step 2: Simplify both sides of the equation. Perform the operations needed on both sides. If the two sides of the equation are equal after this step, the value is a solution.

In our exercise, when we substituted \( t = -2 \) into \( \sqrt{t+6} = t \), it turned into \( \sqrt{-2 + 6} = -2 \), which simplifies to \( \sqrt{4} = -2 \). Since \( \sqrt{4} = 2 \), and 2 does not equal -2, so \( t = -2 \) is not a solution. However, substituting \( t = 3 \), we get \( \sqrt{3+6} = 3 \) simplifying to \( \sqrt{9} = 3 \), which is true because \( \sqrt{9} = 3 \). Therefore, \( t = 3 \) is a solution.

Elementary Algebra

Elementary algebra involves the basic building blocks of algebra which include operations with numbers and variables, solving equations, and understanding functions. Let's connect this to our problem:

• The equation \( \sqrt{t+6} = t \) involves both a variable and a square root function.
• To solve it, basic algebraic skills are necessary to rearrange and simplify expressions.
• Substituting values and simplifying expressions are part of elementary algebra techniques.
• Understanding how to isolate the variable in an equation is key.

In our exercise, we used substitution (replacing \( t \) with given values) and simplification (solving the square root). Both these steps are fundamental aspects of elementary algebra.