Calculus-Based Physics by Jeffrey W. Schnick - HTML preview

Chapter 1 Mathematical Prelude

Example 1-2: Quadratic Formula Example Problem

Given

24

3 + x =

(1-10)

x + 1

find x.

At first glance, this one doesn’t look like a quadratic equation, but as we begin isolating x, as we

always strive to do in solving for x, (hey, once we have x all by itself on the left side of the

equation, with no x on the right side of the equation, we have indeed solved for x—that’s what it

means to solve for x) we quickly find that it is a quadratic equation.

Whenever we have the unknown in the denominator of a fraction, the first step in isolating that

unknown is to multiply both sides of the equation by the denominator. In the case at hand, this

yields

(x + )

1 3

( + x) = 24

Multiplying through on the left we find

3x + 3

2

+ x + x = 24

At this point it is pretty clear that we are dealing with a quadratic equation so our goal becomes

getting it into the standard form of the quadratic equation, the form of equation 1-8, namely:

2

ax + bx + c = 0 . Combining the terms involving x on the left and rearranging we obtain

2

x + 4x + 3 = 24

Subtracting 24 from both sides yields

2

x + 4x − 21 = 0

which is indeed in the standard quadratic equation form. Now we just have to use inspection to

identify which values in our given equation are the a, b, and c that appear in the standard

quadratic equation (equation 1-8)

2

ax + bx + c = 0 . Although it is not written, the constant

multiplying the x2, in the case at hand, is just 1. So we have a = 1, b = 4, and c = −21.

Substituting these values into the quadratic formula (equation 1-9):

− b ± b2 − 4ac

x =

a

2

yields

− 4 ± 42 − 4 )(

1

(

− )

21

x =

2 )

1

(

which results in

x = 3 , x = 7

−

8