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15.7 Essential Skills

Topic

  • The quadratic formula

Previous Essential Skills sections introduced many of the mathematical operations you need to solve chemical problems. We now introduce the quadratic formula, a mathematical relationship involving sums of powers in a single variable that you will need to apply to solve some of the problems in this chapter.

The Quadratic Formula

Mathematical expressions that involve a sum of powers in one or more variables (e.g., x) multiplied by coefficients (such as a) are called polynomials. Polynomials of a single variable have the general form

anxn + ··· + a2x2 + a1x + a0

The highest power to which the variable in a polynomial is raised is called its order. Thus the polynomial shown here is of the nth order. For example, if n were 3, the polynomial would be third order.

A quadratic equation is a second-order polynomial equation in a single variable x:

ax2 + bx + c = 0

According to the fundamental theorem of algebra, a second-order polynomial equation has two solutions—called roots—that can be found using a method called completing the square. In this method, we solve for x by first adding −c to both sides of the quadratic equation and then divide both sides by a:

x 2 + b a x = − c a

We can convert the left side of this equation to a perfect square by adding b2/4a2, which is equal to (b/2a)2:

Left side: x 2 + b a x + b 2 4 a 2 = ( x + b 2 a ) 2

Having added a value to the left side, we must now add that same value, b2 ⁄ 4a2, to the right side:

( x + b 2 a ) 2 = − c a + b 2 4 a 2

The common denominator on the right side is 4a2. Rearranging the right side, we obtain the following:

( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2

Taking the square root of both sides and solving for x,

x + b 2 a = ± b 2 − 4 a c 2 a x = − b ± b 2 − 4 a c 2 a

This equation, known as the quadratic formula, has two roots:

x = − b + b 2 − 4 a c 2 a and x = − b − b 2 − 4 a c 2 a

Thus we can obtain the solutions to a quadratic equation by substituting the values of the coefficients (a, b, c) into the quadratic formula.

When you apply the quadratic formula to obtain solutions to a quadratic equation, it is important to remember that one of the two solutions may not make sense or neither may make sense. There may be times, for example, when a negative solution is not reasonable or when both solutions require that a square root be taken of a negative number. In such cases, we simply discard any solution that is unreasonable and only report a solution that is reasonable. Skill Builder ES1 gives you practice using the quadratic formula.

Skill Builder ES1

Use the quadratic formula to solve for x in each equation. Report your answers to three significant figures.

  1. x2 + 8x − 5 = 0
  2. 2x2 − 6x + 3 = 0
  3. 3x2 − 5x − 4 = 6
  4. 2x(−x + 2) + 1 = 0
  5. 3x(2x + 1) − 4 = 5

Solution:

  1. x=−8+82−4(1)(−5)2(1)=0.583 and x=−8−82−4(1)(−5)2(1)=−8.58
  2. x=−(−6)+(−62)−4(2)(3)2(2)=2.37 and x=−(−6)−(−62)−4(2)(3)2(2)=0.634
  3. x=−(−5)+(−52)−4(3)(−10)2(3)=2.84 and x=−(−5)−(−52)−4(3)(−10)2(3)=−1.17
  4. x=−4+42−4(−2)(1)2(−2)=−0.225 and x=−4−42−4(−2)(1)2((−2))=2.22
  5. x=−1+12−4(2)(−3)2(2)=1.00 and x=−1−12−4(2)(−3)2(2)=1.50