


Step 1 : Enter a quadratic function in terms of x.
Step 2 : Choose a command relating to the function f(x) you entered above.
Quadratic EquationsSolving equations is the central theme of algebra. All skills learned lead eventually to the ability to solve equations and simplify the solutions. In previous chapters we have solved equations of the first degree. You now have the necessary skills to solve equations of the second degree, which are known as quadratic equations. QUADRATICS SOLVED BY FACTORINGOBJECTIVESUpon completing this section you should be able to:
A quadratic equation is a polynomial equation that contains the second degree, but no higher degree, of the variable. The standard form of a quadratic equation is ax^{2} + bx + c = 0 when a ≠ 0 and a, b, and c are real numbers. All quadratic equations can be put in standard form, and any equation that can be put in standard form is a quadratic equation. In other words, the standard form represents all quadratic equations. The solution to an equation is sometimes referred to as the root of the equation.
An important theorem, which cannot be proved at the level of this text, states "Every polynomial equation of degree n has exactly n roots." Using this fact tells us that quadratic equations will always have two solutions. It is possible that the two solutions are equal.
The simplest method of solving quadratics is by factoring. This method cannot always be used, because not all polynomials are factorable, but it is used whenever factoring is possible. The method of solving by factoring is based on a simple theorem. If AB = 0, then either A = 0 or B = 0.
We will not attempt to prove this theorem but note carefully what it states. We can never multiply two numbers and obtain an answer of zero unless at least one of the numbers is zero. Of course, both of the numbers can be zero since (0)(0) = 0. Solution Step 1 Put the equation in standard form.
Step 2 Factor completely.
Step 3 Set each factor equal to zero and solve for x. Since we have (x  6)(x + 1) = 0, we know that x  6 = 0 or x + 1 = 0, in which case x = 6 or x =  1.
Step 4 Check the solution in the original equation. If x = 6, then x^{2}  5x = 6 becomes
Therefore, x = 6 is a solution. If x =  1, then x^{2}  5x = 6 becomes Therefore,  1 is a solution. The solutions can be indicated either by writing x = 6 and x =  1 or by using set notation and writing {6,  1}, which we read "the solution set for x is 6 and  1." In this text we will use set notation.
Check the solutions in the original equation.
INCOMPLETE QUADRATICSOBJECTIVESUpon completing this section you should be able to:
If, when an equation is placed in standard form ax^{2} + bx + c = 0, either b = 0 or c = 0, the equation is an incomplete quadratic. Example 1 5x^{2}  10 = 0 is an incomplete quadratic, since the middle term is missing and therefore b = 0. When you encounter an incomplete quadratic with c  0 (third term missing), it can still be solved by factoring.
Notice that if the c term is missing, you can always factor x from the other terms. This means that in all such equations, zero will be one of the solutions. Example 3 Solve for x if x^{2}  12 = 0. Solution Since x^{2}  12 has no common factor and is not the difference of squares, it cannot be factored into rational factors. But, from previous observations, we have the following theorem.
Using this theorem, we have
Note that in this example we have the square of a number equal to a negative number. This can never be true in the real number system and, therefore, we have no real solution. COMPLETING THE SQUAREOBJECTIVESUpon completing this section you should be able to:
From your experience in factoring you already realize that not all polynomials are factorable. Therefore, we need a method for solving quadratics that are not factorable. The method needed is called "completing the square." First let us review the meaning of "perfect square trinomial." When we square a binomial we obtain a perfect square trinomial. The general form is (a + b)^{2} = a^{2} + 2ab + b^{2}.
From the general form and these examples we can make the following observations concerning a perfect square trinomial.
The 7 term immediately says this cannot be a perfect square trinomial. The task in completing the square is to find a number to replace the 7 such that there will be a perfect square. Consider this problem: Fill in the blank so that "x^{2} + 6x + _______" will be a perfect square trinomial. From the two conditions for a perfect square trinomial we know that the blank must contain a perfect square and that 6x must be twice the product of the square root of x^{2} and the number in the blank. Since x is already present in 6x and is a square root of x^{2}, then 6 must be twice the square root of the number we place in the blank. In other words, if we first take half of 6 and then square that result, we will obtain the necessary number for the blank. Therefore x^{2} + 6x + 9 is a perfect square trinomial. Now let's consider how we can use completing the square to solve quadratic equations. Example 5 Solve x^{2} + 6x  7 = 0 by completing the square.
Solution First we notice that the 7 term must be replaced if we are to have a perfect square trinomial, so we will rewrite the equation, leaving a blank for the needed number. At this point, be careful not to violate any rules of algebra. For instance, note that the second form came from adding +7 to both sides of the equation. Never add something to one side without adding the same thing to the other side. Now we find half of 6 = 3 and 3^{2} = 9, to give us the number for the blank. Again, if we place a 9 in the blank we must also add 9 to the right side as well.
Now factor the perfect square trinomial, which gives
Example 6 Solve 2x^{2} + 12x  4 = 0 by completing the square. Solution This problem brings in another difficulty. The first term, 2x^{2}, is not a perfect square.
We now add 2 to both sides, giving
Example 7 Solve 3x^{2} + 7x  9 = 0 by completing the square. Solution Step 1 Divide all terms by 3.
Step 2 Rewrite the equation, leaving a blank for the term necessary to complete the square. Step 3 Find the square of half of the coefficient of x and add to both sides.
Step 4 Factor the completed square. The factoring should never be a problem since we know we have a perfect square trinomial, which means we find the square roots of the first and third terms and use the sign of the middle term. Step 5 Take the square root of each side of the equation. Step 6 Solve for x (two values).
Follow the steps in the previous computation and then note especially the last ine. What is the conclusion when the square of a quantity is equal to a negative number? "No real solution."
In summary, to solve a quadratic equation by completing the square, follow this stepbystep method. Step 1 If the coefficient of x2 is not 1, divide all terms by that coefficient.
THE QUADRATIC FORMULAOBJECTIVESUpon completing this section you should be able to:
The standard form of a quadratic equation is ax^{2} + bx + c = 0. This means that every quadratic equation can be put in this form. In a sense then ax^{2} + bx + c = 0 represents all quadratics. If you can solve this equation, you will have the solution to all quadratic equations. We will solve the general quadratic equation by the method of completing the square.
This form is called the quadratic formula and represents the solution to all quadratic equations.
To use the quadratic formula you must identify a, b, and c. To do this the given equation must always be placed in standard form.
Not every quadratic equation will have a real solution.
There is no real solution since 47 has no real square root.
This solution should now be simplified. WORD PROBLEMSOBJECTIVESUpon completing this section you should be able to:
Certain types of word problems can be solved by quadratic equations. The process of outlining and setting up the problem is the same as taught in chapter 5, but with problems solved by quadratics you must be very careful to check the solutions in the problem itself. The physical restrictions within the problem can eliminate one or both of the solutions. Example 1 If the length of a rectangle is 1 unit more than twice the width, and the area is 55 square units, find the length and width. Solution The formula for the area of a rectangle is Area = Length X Width. Let x = width, 2x + 1 = length.
At this point, you can see that the solution x = 11/2 is not valid since x represents a measurement of the width and negative numbers are not used for such measurements. Therefore, the solution is width = x = 5, length = 2x + 1 = 11.
Both solutions check. Therefore, the solution set is .
Example 3 If a certain integer is subtracted from 6 times its square, the result is 15. Find the integer. Solution Let x = the integer. Then Since neither solution is an integer, the problem has no solution.
Example 4 A farm manager has 200 meters of fence on hand and wishes to enclose a rectangular field so that it will contain 2,400 square meters in area. What should the dimensions of the field be? Solution Here there are two formulas involved. P = 2l + 2w for the perimeter and A = lw for the area.
We can now use the formula A = lw and substitute (100  l) for w, giving The field must be 40 meters wide by 60 meters long.
Note that in this problem we actually use a system of equations P = 2 l + 2 w In general, a system of equations in which a quadratic is involved will be solved by the substitution method. (See chapter 6.) SUMMARYKey Words
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