Problem 17 - Algebra and Trigonometry

Question

Solve the equation by completing the square. $$ x^{2}+2 x-5=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = -1 + \sqrt{6} \) and \( x = -1 - \sqrt{6} \).

Step 1: Move Constants to the Right Side

Start by moving the constant term to the right side of the equation. This will help in isolating the quadratic and linear terms on the left.\[ x^2 + 2x = 5 \]

Step 2: Find the Square Completing Term

To complete the square, calculate \( \left(\frac{b}{2}\right)^2 \), where \( b \) is the coefficient of \( x \). Here \( b = 2 \).\[ \left(\frac{2}{2}\right)^2 = 1 \]

Step 3: Add the Square Completing Term

Add this square completing term to both sides of the equation to complete the square.\[ x^2 + 2x + 1 = 5 + 1 \]

Step 4: Write the Left Side as a Binomial Square

Now, the left side of the equation forms a perfect square trinomial. Rewrite it as a binomial square.\[ (x + 1)^2 = 6 \]

Step 5: Solve for x by Taking the Square Root

Take the square root of both sides of the equation to solve for \( x \). Remember to consider both positive and negative roots.\[ x + 1 = \pm \sqrt{6} \]

Step 6: Isolate x

Finally, solve for \( x \) by moving the constant term on the left side of the equation to the right.\[ x = -1 \pm \sqrt{6} \]

Key Concepts

Quadratic Equations

A quadratic equation is any equation that has the standard form: \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The highest power of the variable \( x \) in a quadratic equation is 2, and that is what defines it as a "quadratic."

Quadratic equations often have two solutions because they are based on a squared component. These solutions can be found using different methods like factoring, using the quadratic formula, or completing the square.

For our exercise, we choose to solve the quadratic equation \( x^2 + 2x - 5 = 0 \) by completing the square. This method is useful because it transforms the equation into a perfect square trinomial that is easy to solve for \( x \).

Binomial Square

When completing the square, the goal is to transform a quadratic equation into a form that includes a binomial square. A binomial square looks like this:\[ (x + p)^2 \]

Completing the square involves adding a particular value to the equation to make it a perfect square trinomial. The specific value is determined using the formula:

• Calculate \( \left(\frac{b}{2}\right)^2 \), where \( b \) is the coefficient of \( x \).
• Add this value to both sides of the equation.

In our example:

• The coefficient \( b \) is 2.
• We calculate \( \left(\frac{2}{2}\right)^2 = 1 \).
• Add \( 1 \) to both sides to form \( x^2 + 2x + 1 = 6 \).

Now, this trinomial \( x^2 + 2x + 1 \) can be expressed as the binomial square \( (x + 1)^2 \), making it simpler to solve.

Solve for x

After converting the equation into a binomial square, the next step is to solve for \( x \). The equation \( (x + 1)^2 = 6 \) is easier to solve by taking the square root of both sides.

Steps to Solve for \( x \):

• Take the square root of both sides: \( x + 1 = \pm \sqrt{6} \). Remember, the square root yields both positive and negative results.
• Solve for \( x \) by isolating it: \( x = -1 \pm \sqrt{6} \).

The solutions are \( x = -1 + \sqrt{6} \) and \( x = -1 - \sqrt{6} \).

This approach clearly shows the transformation from a complicated quadratic equation to a straightforward linear equation, making it easier to pinpoint the solutions for \( x \).