Problem 30 - Intermediate Algebra

Question

Use the method of completing the square to solve each quadratic equation. $$ x^{2}+7 x+2=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = -\frac{7}{2} + \frac{\sqrt{41}}{2} \) and \( x = -\frac{7}{2} - \frac{\sqrt{41}}{2} \).

Step 1: Rearrange the Equation

Make sure the quadratic equation is in the standard form: \[ x^2 + 7x + 2 = 0 \] This equation is already in the correct form.

Step 2: Identify Coefficients

Identify the coefficients in the equation: \[ a = 1, \ b = 7, \ c = 2 \] We are interested in the \(b\) term for completing the square.

Step 3: Calculate the Term for Completing the Square

To complete the square, take half of the \( b \) coefficient, \(7\), and then square it. \[ \left( \frac{7}{2} \right)^2 = \frac{49}{4} \] This value will be used to complete the square.

Step 4: Add and Subtract the Calculated Value

Add and subtract \( \frac{49}{4} \) inside the equation to maintain the equality:\[ x^2 + 7x + \frac{49}{4} - \frac{49}{4} + 2 = 0 \] This creates a perfect square trinomial within the equation.

Step 5: Rewrite as a Perfect Square

Rewrite the equation:\[ \left( x + \frac{7}{2} \right)^2 - \frac{49}{4} + 2 = 0 \]

Step 6: Simplify the Equation

Combine the constant terms:\[ \left( x + \frac{7}{2} \right)^2 = \frac{49}{4} - 2 \] \[ \frac{49}{4} - \frac{8}{4} = \frac{41}{4} \] The equation simplifies to:\[ \left( x + \frac{7}{2} \right)^2 = \frac{41}{4} \]

Step 7: Solve for x by Taking the Square Root

Apply the square root to both sides of the equation:\[ x + \frac{7}{2} = \pm \sqrt{\frac{41}{4}} \] \[ x + \frac{7}{2} = \pm \frac{\sqrt{41}}{2} \]

Step 8: Isolate x

Subtract \( \frac{7}{2} \) from both sides to solve for \( x \):\[ x = -\frac{7}{2} \pm \frac{\sqrt{41}}{2} \] This results in two possible solutions:\[ x = -\frac{7}{2} + \frac{\sqrt{41}}{2} \] \[ x = -\frac{7}{2} - \frac{\sqrt{41}}{2} \]

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \) because if \( a \) equals zero, the equation becomes linear, not quadratic. Quadratics are prominent in algebra due to their distinct U-shaped graph known as a parabola.
Parabolas can open upwards or downwards depending on the sign of the coefficient \( a \):

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

Solving quadratic equations can be approached in several ways, such as factoring, completing the square, or using the quadratic formula. Each method has its own benefits and ideal cases based on the equation's structure and the given coefficients.

Perfect Square Trinomial

A perfect square trinomial is a special form of a quadratic expression that can be written as the square of a binomial. It has the format \( (x + d)^2 \), which expands to \( x^2 + 2dx + d^2 \) upon multiplication. Recognizing a perfect square trinomial simplifies many algebraic processes, such as solving equations and factoring.
To create a perfect square trinomial from a typical quadratic equation \( ax^2 + bx + c \):

• Focus on the \( b \) term, which represents \( 2d \).
• Divide \( b \) by 2 to find \( d \), then square it to compute \( d^2 \).

Once you calculate \( d^2 \), add and subtract this value within the equation. Completing the square using this method allows algebraic manipulation into simple binomial form, making it easier to solve using square roots.

Square Root Method

The square root method is a straightforward technique that can be used to solve quadratic equations, especially after a quadratic has been transformed into a perfect square trinomial form. This involves isolating the squared term and then applying the square root operation to both sides of the equation.
The process includes:

• Convert the equation into a form where the left-hand side is a perfect square trinomial.
• Isolate the square by rearranging any constants to the other side.
• Apply the square root to both sides: remember this introduces \( \pm \) because both positive and negative values are valid roots.

For example, if we have \( (x + \frac{7}{2})^2 = \frac{41}{4} \), taking the square root will help solve for \( x \): \( x + \frac{7}{2} = \pm \frac{\sqrt{41}}{2} \). Finally, continue to solve for \( x \) by isolating it through further algebraic steps. This method is efficient and less prone to error when the quadratic is neatly expressed as a perfect square.