# 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 :

(variable-stuff)^{2} = (a number).

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

**Example.** Solve the equation for x.

(a) x^{2} − 49 = 0.

x^{2} − 49 = 0

x^{2} = 49

x = ±7

(b) x^{2} + 16 = 0.

(c) (x − 5)^{2} = 4.

(x − 5)^{2} = 4

x − 5 = ±2

x − 5 = 2 gives x = 7 and x − 5 = −2 gives x = 3. The solutions are x = 7 and x = 3.

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 x^{2} + ax.

What would you need to add to x ^{2} + ax to make it a perfect square? If
it comes from (x + c)^{2} =

x^{2} + 2cx + c^{2}, then the middle (x) terms must be the same.
So

2cx = a, c =a/2.

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 x^{2} + 6x = 16 by
completing the square.

The x term is 6x. Half of 6 is 3, and 3^{2} = 9. Therefore, I should add
9 to both sides:

x^{2} + 6x = 16

x^{2} + 6x + 9 = 25

(x + 3)^{2} = 25

x + 3 = ±5

x + 3 = 5 gives x = 2; x + 3 = −5 gives x = −8. The
solutions are x = 2 and x = −8.

**Example.** Solve x^{2} − 10x = 16 by completing the square.

The x term is −10. Half of −10 is −5, and (−5)^{2} = 25. Therefore, I
should add 25 to both sides:

gives
gives The solutions are

As an application of completing the square, I’ll look at equations of circles.
The standard form for the

equation of a circle with radius r and center (a, b) is

(x − a)^{2} + (y − b)^{2} = r^{2}.

**Example.** (a) Find the center and radius of the
circle whose equation is

(x − 5)^{2} + (y + 3)^{2} = 36.

The center is (5,−3) and the radius is 6.

(b) What is the equation of the circle whose center is
(−4, 8) and whose radius is 1.1?

(x − (−4)^{2} + (y − 8)^{2} = 1.1^{2}, or (x + 4)^{2}
+ (y − 8)^{2} = 1.21.

(Don’t multiply out the left side — leave it as is.)

You can find the center and radius of a circle whose
equation is not in standard form by completing the

square.

**Example**. Find the center and radius of the circle

x^{2} − 2x + y^{2} + 6x = 15,

To complete the square in x, I have −2/2= −1, and (−1)^{2}
= 1. To complete the square in y, I have

6/2= 3, and 3^{2} = 9. So I have

x^{2} − 2x + 1 + y^{2} + 6x + 9 = 15 + 1 + 9, or (x − 1)^{2}
+ (y + 3)^{2} = 25.

The center is (1,−3) and the radius is

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

ax^{2} + bx + c = 0.

(If a = 0, 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 (x^{2}) 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 b/2a , its
area is

b^{2}/4a^{2} . 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 ax^{2}
+ bx + c = 0 are

Example. Use the Quadratic Formula to solve x^{2}
− 4x + 2 = 0.

The roots are x = 2 + and x = 2 −
.

Example. Use the Quadratic Formula to solve 2x^{2}
− 8x + 5 = 0.

The roots are and

Example. Use the Quadratic Formula to solve x^{2}
+ 2x + 5 = 0.

Example. Use the Quadratic Formula to solve x^{3}
− 1 = 0.

x^{3} −1 = (x−1)(x^{2} +x+1), so x = 1 is
a root. To find the roots for x^{2} +x+1 = 0, I use the Quadratic

Formula:

The roots are x = 1 and

The Root Theorem says that if p(x) is a polynomial , then x
− c is a factor of p (x) if and only c is a

root of p(x).

To say that x−c is a factor of p(x) is the same as saying
that x−c divides p(x) (“evenly”). For example,

x − 3 divides x^{2} + 4x − 21, because

x^{2} + 4x − 21 = (x − 3)(x + 7).

Example. Factor 3x^{2} − 16x − 35.

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

Now

The root x = 7 corresponds to a factor x − 7; the root x =
−5/3

corresponds to a factor x +5/3. But would have
x^{2} as its square term, whereas the original quadratic has 3x^{2}.
Therefore, I have to multiply by 3 to get the correct factorization:

Example. Factor x^{2} − 6x + 6.

Factoring is equivalent to find the roots of x^{2} − 6x + 6 = 0.

The roots are x = 3 + and x = 3 − . 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.

c = 4 corresponds to the factor x−4. c = −7 corresponds to the factor x−(−7) = x+7. The quadratic

(x − 4)(x + 7) = x^{2} + 3x − 28

has roots 4 and −7.

Example. Find a quadratic equation whose roots are 2 −
and 2 + .

c = 2 − corresponds to the factor x − (2 −
). c = 2 +
corresponds to the factor x − (2 + ).

The quadratic

has roots 2 − and 2 + .

The term b^{2} − 4ac in the quadratic formula is
called the discriminant; it tells you about the roots of

the quadratic equation.

(a) If b^{2} − 4ac > 0, there are two (different) real roots.

(b) If b^{2} − 4ac = 0, there is one real root. (This is called a double
root or a repeated root.)

(c) If b^{2} − 4ac < 0, there are two complex roots. The complex roots
have the form p ± qi.

Example. For what value or values of k does

2x^{2} + kx + 5 = 0

have exactly one root?

The discriminant is

b^{2} − 4ac = k^{2} − 4(2)(5) = k^{2} − 40.

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

x^{2} − 4x + (5 + k^{2}) = 0

has complex roots .

The discriminant is

b^{2} − 4ac = 16 − 4(5 + k^{2}) = −4 − 4k^{2}.

Since −4 − 4k^{2} is negative no matter what k is, the equation always
has complex roots.

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