Problem 91 - College Algebra

Question

Solve each equation in Exercises 73-98 by the method of your choice. \(x^{2}=4 x-7\)

Step-by-Step Solution

Answer

The solutions for the equation \(x^{2} - 4x + 7 = 0\) are \(2 + i\sqrt{3}\) and \(2 - i\sqrt{3}\).

Step 1: Organize in Quadratic Form

Rearrange the equation in the standard quadratic form. Subtract \(4x\) from both sides to move it to the left with the other \(x\) term: \(x^{2} -4x= -7\). Then, add 7 to both sides of the equation to get \( x^{2} - 4x +7 = 0 \).

Step 2: Solve the Equation

The equation \(x^{2} -4x +7 = 0\) cannot be factored nicely so instead we have to use the quadratic formula which is \( x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \). For the equation \( x^{2} -4x +7 = 0 \), a = 1, b = -4, and c = 7. Substituting these values into the quadratic formula will give us the solutions for x.

Step 3: Use the Quadratic Formula

Substitute the values a = 1, b = -4, and c = 7 into the quadratic formula. \( x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(7)}}{2(1)} \) which simplifies to \( x = \frac{4 \pm \sqrt{16-28}}{2} = \frac{4 \pm \sqrt{-12}}{2} \). Since the discriminant is negative, there are no real number solutions, only complex ones.

Step 4: Simplify the Result

We can rewrite \(\sqrt{-12}\) to \(2i\sqrt{3}\) using the property \(\sqrt{-1} = i\). This gets us \( x = \frac{4 \pm 2i\sqrt{3}}{2} = 2 \pm i\sqrt{3}\). So, the solutions are \(2 + i\sqrt{3}\) and \(2 - i\sqrt{3}\).

Key Concepts

Quadratic Formula

Solving quadratic equations can be tricky, especially when they don't factor nicely. This is where the quadratic formula comes in handy. The **quadratic formula** provides a step-by-step approach to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.
• The "\( \pm \)" symbol indicates two possible solutions: one by adding, the other by subtracting.

By plugging the appropriate values of \( a \), \( b \), and \( c \) into this formula, you can find the solutions to any quadratic equation. This method works even when numbers are not easily factorable or when dealing with non-real numbers.

Complex Numbers

When we encounter a negative number inside the square root in our quadratic formula, we delve into the world of **complex numbers**. Unlike real numbers, complex numbers combine both real and imaginary components. An imaginary number is a multiple of the imaginary unit \( i \), which is defined as \( \sqrt{-1} \).
A complex number is expressed as \( a + bi \), where:

• \( a \) is the real part.
• \( b \) is the imaginary part.
• The imaginary unit \( i \) satisfies \( i^2 = -1 \).

In our example, we ended up with \( \sqrt{-12} \), which besides being negative, is simplified to \( 2i\sqrt{3} \). This leads us to solutions like \( 2 + i\sqrt{3} \) and \( 2 - i\sqrt{3} \), which are complex numbers showing both their real and imaginary parts. Complex numbers are crucial in many mathematical fields and have practical applications in engineering and physics.

Discriminant

One key part of the quadratic formula is the **discriminant**, represented by \( b^2 - 4ac \). It's vital in determining the nature of the roots of the quadratic equation before you even solve it completely.
The value of the discriminant tells us:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution (a repeated root).
• If \( b^2 - 4ac < 0 \), we end up with complex solutions as seen in the exercise.

In the exercise provided, \( b^2 - 4ac = -12 \), a negative discriminant. This signifies that the roots of the equation are not real numbers, but complex. Knowing the discriminant can save time by indicating the type of solutions before going through the entire formula process.

Standard Form of a Quadratic Equation

Quadratic equations need to be in a specific format for the quadratic formula and many solving methods to work. This format is known as the **standard form of a quadratic equation**, expressed as:
\[ax^2 + bx + c = 0\]In this setup, \( a \), \( b \), and \( c \) are the constants and \( x \) is the variable.

• \( a eq 0 \), as it ensures the equation remains quadratic and doesn't simplify into a linear equation.
• All terms must be on one side of the equation, with zero on the other side, to apply solving techniques like the quadratic formula.

In the original exercise, we rearranged \( x^2 = 4x - 7 \) by moving everything to one side: \( x^2 - 4x + 7 = 0 \). Putting the equation in this form makes it easier to identify coefficients and use efficient solving strategies.