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(1)The product of two consecutive positive odd number is 195. By constructing a quadratic equation and solving it, find the two numbers.

(2)A Student bought some packets of pens for #2160. If she had paid #24 less for each packet,she could have bought three more packets. How many packets did she buy?

(3)find the positive number n such that twice its square is equal to six times the number?

(4)find the number n such that when 1/3 of it is added to 8, the result is the same as when 1/2 of it is subtracted from 18.

Let n be a positive integer, then 2n+1 is an odd number, and 2n+3 is the next odd number.

"The product of two consecutive positive odd number is 195."

(2n+1)(2n+3) = 195

4n² + 8n + 3 = 195

4n² + 8n - 192 = 0

n² + 2n - 48 = 0

Solve by factoring:

n² + 2n - 48 = (n+8)(n-6)

n = -8, 6 However, n must be positive, so n = 6.

2n+1 = 13

2n+3 = 15

The two consecutive odd numbers are 13 and 15.

:::::

"A student bought some packets of pens for #2160."

If x is the cost of one packet, then 2160/x is the number of packets bought.

"If she had paid #24 less for each packet, she could have bought three more packets."

2160/(x-24) = 3 + 2160/x = (3x+2160)/x

720/(x-24) = (x+720)/x

720x = (x+720)(x-24) = x² + 720x - 24x - 17280

x² - 24x - 17280 = 0

Solve using the Quadratic Formula:

x = [24 ± √(24² + 4·1·17280)]/(2·1)

= [24 ± √69696]/2

= [24 ± 264]/2

= -120, 144

Since the price cannot be negative,

x = 144

2160/x = 15

She bought 15 packets of pens.

:::::

"find the positive number n such that twice its square is equal to six times the number"

2n² = 6n

n² = 3n

n = 3

Check:

2·3² = 18

6·3 = 18

:::::

"find the number n such that when 1/3 of it is added to 8, the result is the same as when 1/2 of it is subtracted from 18."

n/3 + 8 = 18 - n/2

n/3 + n/2 = 10

2n/6 + 3n/6 = 10

5n/6 = 10

n = 10(6/5) = 12

Check:

12/3 + 8 = 12

18 - 12/2 = 12

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