# Referring to what you have learned, you will solve this inequality: (x + 3)(x – 4) < 7. To solve (x + 3)(x – 4) < 7, start by setting (x + 3) = 7 and (x – 4) = 7. Now solve the two...

- Referring to what you have learned, you will solve this inequality:

(x + 3)(x – 4) < 7. - To solve (x + 3)(x – 4) < 7, start by setting (x + 3) = 7 and (x – 4) = 7.
- Now solve the two equations.
- Provide a detailed explanation of why using this method to solve the inequality does not give the correct answer.
- To show your work, attach a MS Word document to your response. Use Microsoft Equation Editor or a similar tool to help you insert mathematical equations and symbols in your document. For assistance, refer to the Microsoft Equation Editor Information document.
- Clearly identify each step in your problem-solving process, demonstrating your progress at each stage.
- Clearly identify your final answer.

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Expert Answers

embizze | Certified Educator

You are asked to solve the inequality (x+3)(x-4)<7:

(1) Suppose that you incorrectly begin by solving x+3=7 and x-4=7. You will get two values for x; x=-3 and x=4.

These you might suppose to be the upper and lower bounds for the inequality. Thus you might get -3<x<4 as an answer. All numbers in this range are indeed solutions, but not every solution is in the interval. For instance, 4.5 is a solution since (4.5+3)(4.5-4)=(7.5)(.5)=3.75<7.

The problem is that there is no 7 product rule for multiplication. If one of the factors is 7, the other need not be 1. This is opposed to the zero product property: if ab=0 then either a=0, b=0 or both a and b are zero.

(2) The correct solution: (x+3)(x-4)<7(x+3)(x-4)-7<0`x^2-x-12-7<0 ``x^2-x-19<0 ` To solve the inequality, set the expression on the left equal to zero. Then test the intervals created by the zeros: `x^2-x-19=0 ==> x=1/2(1 pm sqrt(77) )`(Use the quadratic formula or complete the square.)For `x<1/2(1-sqrt(77)) ` the function is greater than zero so not a solution. For `1/2(1-sqrt(77))<x<1/2(1+sqrt(77)) ` the function is less than zero so solutions. For `x>1/2(1+sqrt(77)) ` the function is positive so not a solution. Geometrically, `f(x)=x^2-x-19 ` is a parabola opening up with x-intercepts at `x=1/2(1 pm sqrt(77)) ` , so the part of the parabola beneath the x-axis is between the intercepts.