Problem 26 - Algebra 1: Concepts and Skills

Question

Solve the equation or write no real solution. Write the solutions as integers if possible. Otherwise, write them as radical expressions. $$ t^{2}=39 $$

Step-by-Step Solution

Answer

The solutions to the equation are \( t = \sqrt{39} \) and \( t = -\sqrt{39} \)

Step 1: Isolate the quadratic part

Start by isolating the part of the equation where \( t \) is squared, which is currently set to equal 39. In this case, the equation is already isolated to \( t^{2}=39 \).

Step 2: Take the square root of both sides

As 't' is square rooted when \( t^{2} = 39 \), now the square root of both sides needs to be taken to make 't' the subject. This gives us \( t = \pm \sqrt{39} \). The \( \pm \) symbol is used because when a numbers square root is taken, it results in two possible values, one positive and one negative.

Step 3: Simplify the solution

As there are no other rational numbers that simplify \( \sqrt{39} \), we can leave the solution in its radical expression form. The final answer will remain as \( t = \pm \sqrt{39} \).

Key Concepts

Radical Expressions

Radical expressions often scare students, but they're just a way to represent roots of numbers. When you see a symbol like \( \sqrt{} \), you're dealing with a radical. In the equation \( t^2 = 39 \), the solution involves taking the square root of 39, which results in a radical expression \( \sqrt{39} \). Radical expressions are useful because they give us a precise way to express numbers that aren't perfect squares—those numbers like 39 in our example.

• Perfect squares are numbers like 1, 4, 9, and so on.
• Non-perfect squares mean we'll get an expression like \( \sqrt{39} \) instead of a neat whole number like 5 or 7.

Often, you can't simplify radical expressions further unless the number under the square root can be factored into smaller squares. With \( 39 \), since there aren't smaller square factors, it stays as \( \sqrt{39} \).

Square Root Method

The square root method is a straightforward way to solve simple quadratic equations, like \( t^2 = 39 \). Here's how it works: isolate the squared term (which we already have in \( t^2 = 39 \)), then take the square root of both sides. This method works because squaring and square rooting are inverse operations—they "undo" each other. Taking the square root of \( t^2 \) simply gives us \( t \). To keep both possibilities open, remember to use the \( \pm \) sign, so your solutions become \( t = \pm \sqrt{39} \).

• Always consider both the positive and negative square roots.
• This rule holds because squaring any positive or negative number gives a positive result.

Using the square root method is neat because it quickly provides the solution without much fuss. Suitably, it's perfect for quadratic equations formatted like this example, where no linear term (terms with just \( t \)) complicate the calculation.

Quadratic Equations

Quadratic equations are a fundamental part of algebra. They often come in the form \( ax^2 + bx + c = 0 \). In our example, however, the equation simplifies to the special case where \( b \) and \( c \) are zero: \( t^2 = 39 \).The algebraic challenge is to find the value of the variable (here, \( t \)) that satisfies the equation. While these equations often have two solutions, our simplified case only requires finding \( \pm \sqrt{39} \).Quadratics can be solved using:

• Factoring (when equations are simple enough).
• The quadratic formula (useful when \( ax^2 + bx + c \) is complex).
• Completing the square (another algebraic method).
• And sometimes, like in our problem, the simple square root method.

Understanding the nature of quadratic equations helps in choosing the best method to reach the solution efficiently.

Integers

Integers are the set of whole numbers, including all positives, negatives, and zero. They don't include fractions or decimals.In the context of solving quadratic equations, solutions can sometimes be integers. You'll end up with integers if the number you're taking the square of is a perfect square—like 4 or 49, where \( \sqrt{4} = 2 \) and \( \sqrt{49} = 7 \).In our example, however, 39 is not a perfect square. Therefore, the solution remains in a radical form (\( t = \pm \sqrt{39} \)).

• Remember: radical solutions and integers are distinct.
• If a solution can't be simplified to an integer, it's typically left as a radical.

In algebra, recognizing when values will yield integers or radicals helps immensely in simplifying solves of quadratic equations.