Problem 32 - Algebra 2

Question

Solve each equation by completing the square. \(2 x^{2}+3 x-5=0\)

Step-by-Step Solution

Answer

The solutions are \( x = 1 \) and \( x = -\frac{5}{2} \).

Step 1: Move the constant term to the other side

Start by moving the constant term \( -5 \) to the other side of the equation. This transforms the equation into the form of \( 2x^2 + 3x = 5 \).

Step 2: Create a perfect square trinomial

Divide all terms by 2 to simplify the equation: \(x^2 + \frac{3}{2}x = \frac{5}{2}\).To complete the square, take half of the coefficient of x, square it, and add it to both sides: \( \left(\frac{3}{4}\right)^2 = \frac{9}{16} \). The equation becomes \( x^2 + \frac{3}{2}x + \frac{9}{16} = \frac{5}{2} + \frac{9}{16} \).

Step 3: Simplify the equation

Rewrite the left side as a perfect square: \( \left(x + \frac{3}{4}\right)^2 \).On the right side, convert \( \frac{5}{2} \) to a common denominator: \( \frac{40}{16} + \frac{9}{16} = \frac{49}{16} \). So the equation is \( \left(x + \frac{3}{4}\right)^2 = \frac{49}{16} \).

Step 4: Solve for x

Take the square root on both sides to find \( x \):\( x + \frac{3}{4} = \pm \frac{7}{4} \).Now solve for \( x \) in each case:1. For \( x + \frac{3}{4} = \frac{7}{4} \), then \( x = 1 \).2. For \( x + \frac{3}{4} = -\frac{7}{4} \), then \( x = -\frac{5}{2} \).

Key Concepts

Quadratic Equation

The foundational idea of a quadratic equation is simple, yet powerful.
A quadratic equation is any equation that can be rearranged into the form of \( ax^2 + bx + c = 0 \), where

• \(a\) is not zero (otherwise it's not quadratic).
• \(b\) and \(c\) can be any real numbers.

The equation given in this exercise, \( 2x^2 + 3x - 5 = 0 \),is a perfect example of a quadratic equation.
Let’s journey through solving it by completing the square. The structure of a quadratic equation guides us through various solving techniques, including completing the square.

Perfect Square Trinomial

A perfect square trinomial arises when a binomial is squared.
This means multiplying the binomial \((x + y)\) by itself to get \((x^2 + 2xy + y^2)\). This expression can be identified or arranged in an equation.
In our exercise, after simplifying, the equation changes to \( x^2 + \frac{3}{2}x = \frac{5}{2} \). To form a perfect square trinomial,

• Take the coefficient of \(x\), which is \(\frac{3}{2}\), and halve it to get \(\frac{3}{4}\).
• Next, square \(\frac{3}{4}\), giving us \(\frac{9}{16}\).
• Add \(\frac{9}{16}\) to both sides to balance the equation.

The new form on the left becomes \( (x + \frac{3}{4})^2 \), a perfect square trinomial.

Solving Equations

Solving equations involves finding the value(s) of the variable that make the equation true.
For a quadratic equation like \( (x + \frac{3}{4})^2 = \frac{49}{16} \),we need to isolate \(x\).

• First, recognize \( (x + \frac{3}{4})^2 \) as a simple expression of the squared form.
• This means you can use the principle of square rooting to solve for \(x\) by removing the squared nature of the equation.

The goal when solving is always to handle operations equally on both sides of the equation, ensuring balance.

Square Root Method

The square root method is a favorite for handling equations of the kind \((x + y)^2 = k\).
This method allows you to "undo" the squaring by taking the square root on both sides.
In our specific example, when you take the square root of both sides, remember that this results in two potential solutions due to the \(\pm\) sign.

• For \( x + \frac{3}{4} = \pm \frac{7}{4} \),solve each scenario separately.
• First equation: \( x + \frac{3}{4} = \frac{7}{4} \), resulting in \( x = 1 \).
• Second equation: \( x + \frac{3}{4} = -\frac{7}{4} \), leading to \( x = -\frac{5}{2} \).

This method systematically leads to the solution of the equation through logical arithmetic steps.