Problem 23 - Algebra and Trigonometry

Question

Solve equation by the square root property. $$ 3(x-4)^{2}=15 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x =\sqrt{5} + 4\) and \(x = -\sqrt{5}+4\).

Step 1: Set the squared term equal to a constant

First, isolate the squared term on one side of the equation. This can be achieved by dividing the entire equation by 3.\[ (x-4)^{2} = \frac{15}{3} = 5 \]

Step 2: Apply the square root property

The next step to undertake is to apply the square root property, which states that if \(a^{2} = b\), then \(a = \sqrt{b}\) or \(a = -\sqrt{b}\). Applying this gives two possible answers:\[ x-4 = \sqrt{5} \]or\[ x-4 = -\sqrt{5} \]

Step 3: Solve for x

Finally, solve for x in each of the two equations derived. This gives:\[ x = \sqrt{5} + 4 \]and\[ x = -\sqrt{5}+4 \].

Key Concepts

Square Root Property

The square root property is a powerful tool in algebra, especially when working with quadratic equations. It helps to simplify equations by removing the square. Basically, if you have an equation of the form \[a^2 = b\]then the solutions for \(a\) are:

• \(a = \sqrt{b}\)
• \(a = -\sqrt{b}\)

This means you take both the positive and negative square roots of \(b\).

Applying this to an example such as \[(x - 4)^2 = 5\]yields two solutions by taking the square roots:

• \(x - 4 = \sqrt{5}\)
• \(x - 4 = -\sqrt{5}\)

This approach helps in finding all possible solutions of a quadratic equation.
Always remember that quadratic equations can have up to two real solutions due to this property.

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. In algebra, symbols, like \(x\) and \(y\), are used to represent numbers and express mathematical relationships.

One key aspect of algebra in solving equations is the concept of balance. What you do to one side of an equation, you must do to the other to keep it balanced. In our example,\[3(x-4)^2=15\], we used algebraic manipulation to isolate the squared term. By dividing both sides by 3, we maintained balance and simplified the equation to:\[(x-4)^2 = 5\].

Practicing algebra helps build a strong foundation for solving more complex mathematical problems. It involves operations such as addition, subtraction, multiplication, and division, in combination with using symbols to solve problems.

Quadratic Expressions

Quadratic expressions are algebraic expressions of degree two. They take the general form:\[ax^2 + bx + c\],where \(a\), \(b\), and \(c\) are constants, and \(x\) is an unknown variable.

In the given problem, the quadratic expression is formed as \[(x-4)^2\].It is a perfect square, which means it has been expressed as a binomial raised to the power of 2. This makes it easier to solve by applying the square root property.

Quadratic expressions often appear in various forms, and solving them requires understanding key algebraic techniques, such as factoring, completing the square, and using the quadratic formula.
These strategies enable you to find the roots or solutions of quadratic equations, which are the values of \(x\) that make the expression equal to zero.