Problem 63 - Algebra 1

Question

Solve the equation or write no solution. Write the solutions as integers if possible. Otherwise write them as radical expressions. $$2 s^{2}-5=27$$

Step-by-Step Solution

Answer

The solutions to the equation are \(s = 4\) and \(s = -4\)

Step 1: Rearrangement

First, rearrange the equation to have it equal to zero: \(2 s^{2}-5-27=0\).

Step 2: Simplification

Simplify the equation by combining like terms: \(2 s^{2}-32=0\).

Step 3: Isolate \(s^{2}\)

To isolate \(s^2\), we divide both sides of the equation by 2: \(s^{2}-16=0\).

Step 4: Use the difference of squares

The equation now is a difference of squares, which can be factored into \((s-4)(s+4)=0\).

Step 5: Solve for s

Finally, we set each factor to zero and solve for \textit{s}, giving \textit{s} = 4 and \textit{s} = -4.

Key Concepts

Factoring

Factoring is a powerful tool for solving quadratic equations. It involves writing an expression as a product of its factors, which are simpler expressions that, when multiplied together, give the original expression. In the given exercise, we have a quadratic equation that can be factored after rearranging and simplifying.

To factor a quadratic expression like the one in the given task, the equation is first put into the form of a quadratic trinomial, but in our exercise, we deal with a difference of squares. This means the expression looks like this:

• \(a^2 - b^2\)

The next step is to recognize patterns or special forms like this, which can then be expressed as:

• \((a-b)(a+b)\)

In the case of the exercise, the terms \((s^2 - 16)\) are recognized as a difference of squares, allowing us to factorize it into:

• \((s-4)(s+4)\)

Once factored, the solutions can easily be found by setting each factor equal to zero. This strategy simplifies the problem significantly and transforms it into a basic arithmetic solution.

Difference of Squares

The "difference of squares" is a specific type of factoring method that applies when you have two perfect squares separated by a subtraction sign. This is identified by an equation of the form \(a^2 - b^2\), where both \(a^2\) and \(b^2\) are perfect squares.

The expression follows a strikingly simple factoring rule that can be memorized to ease solving quadratic problems. The rule is represented as:

• \(a^2 - b^2 = (a-b)(a+b)\)

In the context of the exercise, the perfect square terms are \(s^2\) and \(16\) (since \(16 = 4^2\)). This makes our expression \(s^2 - 16\) a classic difference of squares.

Applying our rule:

• \(s^2 - 16 = (s-4)(s+4)\)

This simple technique turns a quadratic equation into two linear equations. It is a quick way to find where each factor equals zero, giving us the roots of the original quadratic equation.

Radical Expressions

Radical expressions often appear in the solutions of quadratic equations, especially when exact integer solutions are not available. A radical expression includes a square root (or another root), written using the radical symbol \(\sqrt{}\).

In some exercises, quadratic equations do not neatly factor into integers, and the roots involve radicals. This can happen when the discriminant (the expression under the square root in the quadratic formula) is not a perfect square, meaning the solutions will not simplify to integers.

Fortunately, in our given exercise, factoring gave us integer solutions, \(s = 4\) and \(s = -4\). However, when tackling problems with non-integer solutions, these roots must be expressed as radicals. For example, if we had a quadratic that resulted in roots like \(\frac{-b \pm \sqrt{D}}{2a}\), where \(D\) is the discriminant, it would involve a radical unless \(\sqrt{D}\) simplifies to an integer.

Understanding radical expressions and knowing how to simplify them is essential for finding exact solutions and for fully grasping the complete set of roots in more complex problems.