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Quadratic Equations and the Quadratic Formula Study Guide

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Updated on Oct 3, 2011

Introduction to Quadratic Equations and the Quadratic Formula

Algebra as far as the quadratic equation and the use of logarithms are often of value in ordinary cases: but all beyond these is but a luxury; a delicious luxury indeed.

—Thomas Jefferson (1743–1826) Third President of the United States

 In this lesson, you'll learn how to solve a quadratic equation using factoring and the quadratic formula.

A quadratic equation is a quadratic expression with an equal sign. Formally, the quadratic equation is described as ax2 + bx + c = 0. a is the coefficient of the x2 term, and it cannot be 0. Otherwise there would be no x2 term (and then the equation would not be quadratic!). b is the coefficient of the x term, and c is the constant.

Most quadratic equations have two solutions, although some have only one. To solve a quadratic equation, the first step is to get the equation in the form ax2 + bx + c = 0. If the equation is equal to some value other than 0, subtract that value from both sides of the equation. That value will be combined with one of the terms on the left side of the equation, leaving the equation in the form ax2 + bx + c = 0.

Once the equation is in that form, we have two ways to find the solutions. The first method is to factor the equation into two binomials. Now you know why we learned how to do that in Lesson 23. We then take the two binomials and set them each equal to 0. Why? An expression is the product of its factors. If a value of x makes one factor equal to 0, then the product of the factors will be equal to 0, because any quantity multiplied by 0 is 0.

The equation x2 + 9x + 18 = 0 is ready to be factored, because it is already in the form ax2 + bx + c = 0. To factor x2 + 9x + 18, we start as always by listing the factors of x2 and the factors of 18:

      x2: –x, x
      18: –18, –9, –6, –3, –2, –1, 1, 2, 3, 6, 9, 18

The middle term, 9x is positive, and the last term, 18, is positive, so we're looking for two positive numbers:

      (x + 1)(x + 18) = x2 + 19x + 18. The middle term is too large.
      (x + 2)(x + 9) = x2 + 11x + 18. The middle term is still too large.
      (x + 3)(x + 6) = x2 + 9x + 18. Got it!

The factors of x2 + 9x + 18 are (x + 3) and (x + 6). The equation x2 + 9x + 18 = 0 holds true when either (x + 3) equals 0 or (x + 6) equals 0. We set each factor equal to 0 and solve for x:

x + 3 = 0 x + 6 = 0
x = –3 x = –6

The roots, or the solutions of x2 + 9x + 18 = 0, are x = –3 and x = –6. The roots of an equation are the values for which f (x) is 0.

That equation had two roots, but if the trinomial in the equation comes from squaring a binomial, then the equation will have only one root.

To solve the equation x2 + 25 = 10x, we must first put the equation in the form ax2 + bx + c = 0. If we subtract 10x from both sides of the equation, we have x2 – 10x + 25 = 0. Now, we are ready to factor. The factors of x2 are –x, x and the factors of 25 are –25, –5, –1, 1, 5, 25. Because the middle term is negative and the last term is positive, we are looking for two negative numbers that multiply to 25 and add to –10. x2 – 10x + 25 factors into (x – 5)(x – 5). Because x2 – 10x + 25 = 0, we set each of its factors equal to 0. However, both factors are the same. They both are (x – 5), so we only need to set one of them equal to 0:

      x – 5 = 0
      x = 5

The trinomial x2 – 10x + 25 came from squaring (x – 5), so the equation has only one root, 5.

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