# The Quadratic Formula

The quadratic formula is a formual for finding the roots of a quadratic equation. It depends on a procedure called completing the square. Here's the idea. Suppose you can do algebra to get your equation to look like this:

Then you can solve for the variable-stuff by taking the square root of both sides.

Example. Solve the equation for x.

(a) .

(b) .

(c) .

gives and gives . The solutions are and .

Suppose you start out with a quadratic equation where you don't have a perfect square on one side. Suppose, for instance, that the variable-stuff is .

What would you need to add to to make it a perfect square? If it comes from , then the middle (x) terms must be the same. So

Therefore, I need to add to get a perfect square. The procedure is: Take half of the coefficient of the middle term ( ), then square it ( ).

Example. Solve by completing the square.

The x term is . Half of 6 is 3, and . Therefore, I should add 9 to both sides:

gives ; gives . The solutions are and .

Example. Solve by completing the square.

The x term is -10. Half of -10 is -5, and . Therefore, I should add 25 to both sides:

gives ; gives . The solutions are .

Now I'll see how to use completing the square to obtain a formula for solving an aribtrary quadratic equation.

Start with a quadratic equation

Divide by a:

(If , the equation wasn't quadratic to begin with.)

Move the constant term to the other side:

The idea is to add something to both sides to make the left side a perfect square.

Represent the left side as the sum of the areas of a square ( ) and a rectangle ( ). Divide the rectangle into two equal pieces.

Move the bottom piece next to the square. The total area is still the same. What is the area of the square that must be added to "complete the square"?

Since the missing square has sides of length , its area is . This is what I need to add to both sides:

So

Take the square root of both sides and solve for x:

Theorem. ( The Quadratic Formula) The solutions to are

Example. Use the Quadratic Formula to solve .

The roots are and .

Example. Use the Quadratic Formula to solve .

The roots are and .

Example. Use the Quadratic Formula to solve .

Example. Use the Quadratic Formula to solve .

, so is a root. To find the roots for , I use the Quadratic Formula:

The roots are and .

The Root Theorem says that if is a polynomial, then is a factor of if and only c is a root of .

To say that is a factor of is the same as saying that divides ("evenly"). For example, divides , because

Example. Factor .

By the Root Theorem, factoring is equivalent to finding roots. By the Quadratic Formula, the roots are

Now

The root corresponds to a factor ; the root corresponds to a factor . But would have as its square term, whereas the original quadratic has . Therefore, I have to multiply by 3 to get the correct factorization:

Example. Factor .

Factoring is equivalent to find the roots of .

The roots are and . This means that

You can also use the Root Theorem to construct quadratics with given roots.

Example. Find a quadratic equation whose roots are 4 and -7.

corresponds to the factor . corresponds to the factor . The quadratic

has roots 4 and -7.

Example. Find a quadratic equation whose roots are and .

corresponds to the factor . corresponds to the factor . The quadratic

has roots and .

The term in the quadratic formula is called the discriminant; it tells you about the roots of the quadratic equation.

(a) If , there are two (different) real roots.

(b) If , there is one real root. (This is called a double root or a repeated root.)

(c) If , there are two complex roots. The complex roots have the form .

Example. For what value or values of k does

have exactly one root?

The discriminant is

There is exactly one root when the discriminant is 0:

There is exactly one root when .

Example. Show that if k is any real number, the equation

has complex roots.

The discriminant is

Since is negative no matter what k is, the equation always has complex roots.

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Copyright 2016 by Bruce Ikenaga