Problem 10 - Elementary Algebra

Question

For the following problems, solve the equations by completing the square. $$ y^{2}-2 y-24=0 $$

Step-by-Step Solution

Answer

Answer: The solutions are $$y = 6$$ and $$y = -4$$.

Step 1: Identify the terms of the equation

In the given equation, $$y^2 - 2y - 24 = 0$$, identify the square term (\(y^2\)), the linear term (-2y), and the constant term (-24).

Step 2: Rewrite the equation in the form of a perfect square trinomial

Add the constant term (24) to both sides of the equation to isolate the square term and the linear term on the left side. $$y^2 - 2y = 24$$ Next, find the number that completes the square for the trinomial \((y^2 - 2y + \Box)\). For this, divide the linear term coefficient (-2) by 2 and square the result: $$\left(\frac{-2}{2}\right)^2 = 1$$ Now add this number (1) to both sides of the equation to complete the square trinomial: $$y^2 - 2y + 1 = 24 + 1$$ So, the equation becomes: $$y^2 - 2y + 1 = 25$$

Step 3: Factor the perfect square trinomial

Factor the left side of the equation into a square binomial, and simplify the right side: $$\left(y - 1\right)^2 = 25$$

Step 4: Apply the square root property

Take the square root of both sides of the equation. Remember to include both positive and negative roots for the right side of the equation: $$\sqrt{\left(y - 1\right)^2} = \pm\sqrt{25}$$ This simplifies to: $$y - 1 = \pm 5$$

Step 5: Solve for 'y'

Add 1 to both sides of the equation to isolate 'y': $$y = 1 \pm 5$$ This yields the two possible solutions for 'y': $$y = 1 + 5 = 6$$ $$y = 1 - 5 = -4$$ So the solutions to the given equation are $$y = 6$$ and $$y = -4$$.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They take the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) is an unknown variable. In these equations, the term \(ax^2\) is the quadratic term, \(bx\) is the linear term, and \(c\) is the constant term. Solving quadratic equations often involves finding the values of \(x\) that satisfy the equation.
Quadratic equations can be solved using various methods, including:

• Factoring
• Using the quadratic formula
• Graphing
• Completing the square

Each method has its unique advantages, and the method of completing the square is particularly useful when the equation is not easily factorable.

Factoring

Factoring is a method of solving quadratic equations by expressing the quadratic expression as a product of two binomials. If a quadratic equation can be factored, it can make finding solutions simpler. The factored form of a quadratic equation \(ax^2 + bx + c = 0\) is \((px + q)(rx + s) = 0\), where \(p\), \(q\), \(r\), and \(s\) are constants that satisfy the original equation.
To solve the factored equation, set each binomial equal to zero:

• \(px + q = 0\)
• \(rx + s = 0\)

Solving these provides the roots of the quadratic equation. This method is effective and time-saving but relies on the quadratic expression being factorable into rational numbers or integers.

Perfect Square Trinomial

A perfect square trinomial is a special form of a quadratic equation that can be written as the square of a binomial. It follows the pattern \((x + a)^2 = x^2 + 2ax + a^2\). Completing the square is a technique used to transform a quadratic equation into a perfect square trinomial.
In the equation \(y^2 - 2y - 24 = 0\), after isolating the quadratic and linear terms, we use the value \(-2\), dividing it by 2 and squaring the result to find \(1\). By adding \(1\) to both sides, the expression \(y^2 - 2y + 1\) becomes a perfect square trinomial, \((y - 1)^2\). This method simplifies solving the equation using the square root property.

Square Root Property

The square root property is a straightforward method used to solve an equation that involves a perfect square trinomial. Once it's in the form \((y - 1)^2 = 25\), the next step is to apply the square root property.
This property suggests that if \(u^2 = d\), then \(u = \pm\sqrt{d}\). Applying this to the equation, take the square roots of both sides to eliminate the square on the left:

• \(\sqrt{(y - 1)^2} = \pm\sqrt{25}\)

Thus, solving for \(y\), we get \(y - 1 = \pm 5\). Finally, isolate \(y\) by adding \(1\) to both sides, leading to two solutions: \(y = 6\) and \(y = -4\). This method is efficient and leads directly to the solutions without the need for complex calculations.