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# Determine a quadratic equation for each of the following problems in the exercise. Then solve the equation.?

Topic: Determine a quadratic equation for each of the following problems in the exercise. Then solve the equation.?
July 18, 2019 / By Primrose
Question: I am solve the equations I just need help setting them up! 3. The square of a positive interger is equal to the sum of the integer and 12. Find the integer. 6. One positive number is three more than twice another. If the product is 27, find the numbers. 10. One number is three less than twice another. The sum of their squares is 74. Find the numbers. 14. Find two consecutive interfere whose product is 110. 15. Find two consecutive positive integers such that the sum of their squares is 85. 18. The product of two consecutive odd integers is 63, find the integers 21. Te length of the rectangle is twice the width, the area is 72 square feet. Find the length and width of the rectangle. 25. The width of a rectangle is 4ft less than the lenght. The area is 117 square feet. Find the lenght and width of the rectangle. 26. The height of a triangle is 4 feet less than the base.the area of the triangle is 16 square feet. Find the length of the base and height of the triangle. 31. The area of a rectangle is 24 square centimeters. If the perimeter is 20 centimeters, find the lenght and width of the rectangle. 34. One formation for a drill team is rectangular. The number of members in each toe exceeds the number of rows by 3. If there is a total of 108 members in the formations, how many rows are there. 37. The width of a rectangle is 5 meters less than the lenght. If 6 meters are added to both the length and width, the new area will be 300 square meters. Fins the dimensions of the original rectangle. 39. A rancher is going to build a cores with 52 yards of fencing. He is planning to use the barn as one side of the corral. If the area is 320 square yards, what are the dimensions. 41. The length of a rectangle is three tomes the width. If the area is 48 square feet find the lenght of the rectangle.

## Best Answers: Determine a quadratic equation for each of the following problems in the exercise. Then solve the equation.?

Mavis | 1 day ago
Not going to go through all of these, but will help you will the basics for several. #3. The square of a positive integer is equal (x^2 =) to the sum of the integer and 12 (x+12). So, x^2 = x + 12 #6. One positive number is three more than twice another. If you call the second one x, the first one is three more than twice that, or 3 more than 2x, or 2x+3. Since their product is 27, that gives you x (2x+3) = 27 #10. Similar to #6. The 2x+3 becomes 2x-3, and you'll add instead of multiplying. Also, note you want to add the sum of the squares, not the numbers themselves. This will give you x^2 + (2x-3)^2 = 72. #14. Two consecutive numbers will have the form x, and x + 1 (since two consecutive numbers are only one apart). So, x(x+1)=110 #15. Like #14, except now you have to square the numbers and add them. (x)^2 + (x+1)^2 = 85 #18. If you have consecutive even or odd numbers, you are looking at numbers that are two apart. Thus, the numbers would be x and (x + 2). Thus, your setup is x(x + 2) = 63. For the remaining problems, remember you dealing with distances, so you need positive answers, and can ignore the negative ones. #21. You need to know that area of a rectangle = length times width, or A= lw. length is twice the width, so set it up like that. Length is twice width --> L = 2w, so A = lw 72 = 2w(w) #25. Similar to above. Width is 4 less than length, so w = l - 4. A = l w 117 = l (l-4) #26. Formula is now A = 1/2 b h. height is 4 less than base, so h= b-4 A = 1/2 b h 16 = 1/2 (b) (b-4) #31. Back to rectangles. Since Area = lw, and the area is 24, you get 24=lw. Perimeter is 20 cm, giving you 2l+2w = 20. I would rewrite the second one like so: 2l = 20 - 2w l = 10 - w and plug into the area formula. 24 = lw 24 = (10-w)w #34. Just an area problem. A = l*w = (w+3)(w). Notice that number of member in each row exceeds the number of rows, so when you get your two measurements, you want the smaller one. #37. w=l-5 If you increase each by 6 meters, then the new length would be l+6, and the new width would be w+6=(l-5)+6=l+1. So solve 300 = (l+6)(l+1) #39. P=2l+2w Perimeter is 52. 52=2l+2w 52-2l=2w 26-l=w Area is 320, so solve: 320 = l (26-l) #41. Last one. Length is three times the width, so length is 3w. A = lw = (3w)(w) So solve 48 = 3w(w) to find the size of the width. But remember you will want the length, which is three times that.
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Originally Answered: how to solve geometric problems (solving Quadratic equation by completing square)?
Originally Answered: how to solve geometric problems (solving Quadratic equation by completing square)?
(x - 2)(x +6) = 308 x^2 +4x - 12 = 308 x^2 +4x -320 = 0 x1 = -2 + sqrt(324) x1 = -2 + 18 = 16 x2 = - 2 -18 = -20 The length cannot be negative Thus x = 20 ist the solution Length : 16 - 2 = 14 Width: 16 + 6 = 22 Test 14 * 22 = 308

Laura
first, factor the quadratic on the left side of this equation, to give you (x-2)(x+5)=0. Using the property of a product whose answer is 0, then it follows that: x - 2 = 0 and/or x + 5 = 0 solving each of these yields x = 2 or x = -5
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Jewel
I'll set up a few for you, but then, you're on your own. 3. x^2 x+12 6. x(2x+3) = 27 10. x^2 + (2x-3)^2 = 74 14. x(x+14) = 110 18. x(x+2) = 63
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