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The name game: Solutions, roots, and zeros in VS .NET
The name game: Solutions, roots, and zeros
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Algebra lets you describe the x-values that you find when an equation is set equal to 0 in several different ways For instance, when (x 3) (x + 4) = 0, you have two: Solutions to the equation, x = 3 and x = 4 Roots of the equations, 3 and 4, because they make the equation true Zeros for the equation (values that make the equation equal to 0) that occur when x = 3 and x = 4 x-intercepts at (3, 0) and ( 4, 0) The descriptions are often used interchangeably, because you determine these values in exactly the same way
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3: Cracking Quadratic Equations
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The techniques you use to solve the inequalities in this section are also applicable for solving higher degree polynomial inequalities and rational inequalities If you can factor a third- or fourth-degree polynomial (see the previous section to get started), you can handily solve an inequality where the polynomial is set less than zero or greater than zero You can also use the sign-line method to look at factors of rational (fractional) expressions For now, however, consider sticking to the quadratic inequalities To solve the inequality x2 x > 12, for example, you need to determine what values of x you can square so that when you subtract the original number, your answer will be bigger than 12 For instance, when x = 5, you get 25 5 = 20 That s certainly bigger than 12, so the number 5 works; x = 5 is a solution How about the number 2 When x = 2, you get 4 2 = 2, which isn t bigger than 12 You can t use x = 2 in the solution Do you then conclude that smaller numbers don t work Not so When you try x = 10, you get 100 + 10 = 110, which is most definitely bigger than 12 You can actually find an infinite amount of numbers that make this inequality a true statement Therefore, you need to solve the inequality by using the steps I outline in the introduction to this section: 1 Subtract 12 from each side of the inequality x2 x > 12 to move all the terms to one side You end up with x2 x 12 > 0 2 Factoring on the left side of the inequality, you get (x 4)(x + 3) > 0 3 Determine that all the zeroes for the inequality are x = 4 and x = 3 4 Put the zeros in order on a number line, shown in the following figure
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5 Create a sign line to show the signs of the different factors in each interval Between 3 and 4, try letting x = 0 (you can use any number between 3 and 4) When x = 0, the factor (x 4) is negative, and the factor (x + 3) is positive Put those signs on the sign line to correspond to the factors Do the same for the interval of numbers to the left of 3 and to the right of 4 (see the following illustration)
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Part I: Homing in on Basic Solutions
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(x 4)(x + 3) x= 5 ( )( ) 3 (x 4)(x + 3) x=0 ( )( + ) 4 (x 4)(x + 3) x = 10 ( + )( + )
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The x values in each interval are really random choices (as you can see from my choice of x = 5 and x = 10) Any number in each of the intervals gives you the same positive or negative value to the factor 6 To determine the solution, look at the signs of the factors; you want the expression to be positive, corresponding to the inequality greater than zero The interval to the left of 3 has a negative times a negative, which is positive So, any number to the left of 3 works You can write that part of the solution as x < 3 or, in interval notation (see 1), ( , 3) The interval to the right of 4 has a positive times a positive, which is positive So, x > 4 is a solution; you can write it as (4, ) The interval between 3 and 4 is always negative; you have a negative times a positive The complete solution lists both intervals that have working values in the inequality The solution of the inequality x2 x > 12, therefore, is x < 3 or x > 4
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