Problem 64 - Algebra A Combined Function

Question

Solve each equation by completing the square. $$ 9 x^{2}-36 x=-40 $$

Step-by-Step Solution

Answer

The solutions are \( x = 2 \pm \frac{2\sqrt{19}}{3} \).

Step 1: Move Constant to Right Side

Start by moving the constant term from the left to the right side of the equation. The equation is \( 9x^2 - 36x = -40 \). Since we want to simplify the quadratic without the constant, we add 40 to both sides of the equation, leading to \( 9x^2 - 36x + 40 = 0 \). However, to prepare for completing the square, isolate the quadratic and linear term: \( 9x^2 - 36x = 40 \).

Step 2: Divide by Coefficient of x²

The coefficient of \( x^2 \) is 9, so divide the entire equation by 9 to make the coefficient 1: \( x^2 - 4x = \frac{40}{9} \). This step simplifies the equation making it easier to complete the square.

Step 3: Complete the Square

To complete the square, take half of the coefficient of \( x \), square it, and add it to both sides. The coefficient of \( x \) is -4; half of -4 is -2 and (-2)^2 = 4. Add 4 to both sides to form a perfect square trinomial: \( x^2 - 4x + 4 = \frac{40}{9} + 4 \).

Step 4: Simplify the Right Side

Simplify the right side of the equation: \( \frac{40}{9} + 4 \) is equivalent to \( \frac{40}{9} + \frac{36}{9} = \frac{76}{9} \). So, the equation becomes \( x^2 - 4x + 4 = \frac{76}{9} \).

Step 5: Write as a Perfect Square

Now, write the left side of the equation as a perfect square: \( (x-2)^2 = \frac{76}{9} \). The expression \( (x-2)^2 \) is derived from \( x^2 - 4x + 4 \), which is a perfect square trinomial.

Step 6: Solve for x

Take the square root of both sides to solve for \( x \): \( x-2 = \pm \sqrt{\frac{76}{9}} \). This gives two possible solutions. Simplify \( \sqrt{\frac{76}{9}} \) to \( \frac{\sqrt{76}}{3} \) and solve for \( x \): \( x = 2 \pm \frac{\sqrt{76}}{3} \).

Step 7: Simplify Final Answer

Further simplify \( \sqrt{76} \): \( \sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19} \). Thus, the solutions are \( x = 2 \pm \frac{2\sqrt{19}}{3} \).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation. They take the form:

• \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).

In our exercise, the equation is \( 9x^2 - 36x + 40 = 0 \), where:

• The coefficient of \( x^2 \) is 9.
• The coefficient of \( x \) is -36.
• The constant term \( c \) is 40 on the right side.

To solve these equations, different methods can be used, such as factoring, using the quadratic formula, or completing the square.
Quadratic equations typically have two solutions because they represent parabolas on a graph, which typically intersect the x-axis at two points. Each solution can be real or imaginary, based on the discriminant \( b^2 - 4ac \). In our exercise, we are focusing on solving this quadratic equation by completing the square, which involves transforming it into a form that can be easily square rooted.

Perfect Square Trinomial

A perfect square trinomial is a polynomial that can be expressed as the square of a binomial. In simpler terms, it's a three-term expression that factors neatly into something squared. The general form is:

• \((x + p)^2 = x^2 + 2px + p^2\)

Translating this to our exercise:

• The expression \(x^2 - 4x + 4\) is a perfect square trinomial.
• It can be rewritten as \((x-2)^2\).

To identify or create a perfect square trinomial, take these steps:

• Take the coefficient of \( x \), divide it by 2, and then square it.
• For \(-4\): Half is \(-2\), squaring gives \(4\), which is added to both sides.

This process allows us to express a quadratic equation in a way that can be factored, setting the stage for taking the square root. It translates complex algebra into simpler arithmetic, making solving equations more straightforward.

Square Root

The square root is a valuable mathematical operation used to "undo" a square (\( x^2 \)). Given a number \( y \), the square root \( \sqrt{y} \) yields "x," such that \( x^2 = y \).In the context of completing the square, once you have written the equation in the form of a perfect square, taking the square root helps in finding the real solutions of the equation.

Here's how to apply it in our problem:

• You have \((x-2)^2 = \frac{76}{9} \).
• To solve for \( x \), take the square root of both sides: \(x-2 = \pm \sqrt{\frac{76}{9}}\).

Notice the symbol \( \pm \): it means there are two values, as both the positive and negative roots are considered. After calculating the square root:

• Simplify \( \sqrt{\frac{76}{9}} \): It becomes \( \frac{\sqrt{76}}{3} \).
• This translates to \( x = 2 \pm \frac{2\sqrt{19}}{3} \).

Taking square roots in quadratic solutions is a critical step, providing the possible values of \( x \) that make the equation true. These values, often causes that solve real-world parabolas on a graph, help understand many phenomena and are a cornerstone of algebra.