Question 60703: Find two consecutive even integers whose product is 288.
A number is increased by 4 times its reciprocal is 20/3. Find the number.
The sum of the squares of two consecutive even integers is 244. Find the integers.
If the product of two consecutive integers is decreased by 20 times the greater integer, the result is 442. Find the integers.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find two consecutive even integers whose product is 288.
x(x+2) = 288
x^2 + 2x - 288 = 0
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A quick way to factor this is to use your calc to find the square root of 288. It's 16.97. you know that the factors are even and differ by 2, one above 16.97 and one below. obviously it's 16 and 18
(x+18)(x-16) = 0
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x = +16 is probably the the solution we want, however -18 would also be solution
-18*-16 = +288 also
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A number is increased by 4 times its reciprocal is 20/3. Find the number.
x + 4(1/x) = 20/3
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Get rid of the denominators, mult eq by 3x and you have:
3x^2 + 3(4) = 20x
3x^2 - 20x + 12 = 0
(3x - 2)(x - 6) = 0
x = + 6 would be the integer solution
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The sum of the squares of two consecutive even integers is 244. Find the integers.
x^2 + (x+2)^2 = 244
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x^2 + x^2 + 4x + 4 = 244
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2x^2 + 4x + 4 - 244 = 0
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2x^2 + 4x - 240 = 0
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Simplify, divide equation by 2 and you have:
x^2 + 2x - 120 = 0
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Easily factors too:
(x-10)(x+12) = 0
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x = +10 and x = -12, either is a solution
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If the product of two consecutive integers is decreased by 20 times the greater integer, the result is 442. Find the integers.
[x(x+1)] - 20(x+1) = 442
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x^2 + x - 20x - 20 = 442
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x^2 - 19x -20 - 442 = 0
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x^2 - 19x - 462 = 0
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Factors to:
(x-33)(x+14) = 0
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x = +33 and x = -14
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Both are solutions (check on your calc):
33*34 - (20*34) =
and
[-14*-13] - (20*-13) =
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Note that -13 is greater than -14 in this context.
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