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...

  1. Referring to what you have learned, you will solve this inequality: 
    (x + 3)(x – 4) < 7.
  2. To solve (x + 3)(x – 4) < 7, start by setting (x + 3) = 7 and (x – 4) = 7.
  3. Now solve the two equations.
  4. Provide a detailed explanation of why using this method to solve the inequality does not give the correct answer.
  5. 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.
  6. Clearly identify each step in your problem-solving process, demonstrating your progress at each stage.
  7. Clearly identify your final answer.

Asked on by cathy-cobb

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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. We can look at the original problem graphically as well -- we are asked to find when the graph of y=(x+3)(x-4) is below the line y=7. (Note that our solution is a transformation of the original problem -- shift everything down 7 units.) The solutions to (x+3)(x-4)<7 are `1/2(1-sqrt(77))<x<1/2(1+sqrt(77)) `

 

 

 

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