Twice the square of an integer is 5 less then 11 times the integer. Find the integer?

Answer 1

Let

x = the Integer

2x² = Twice the square of the integer

11x = 11 times the integer

-5 = 5 less

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The equation

2x² = 11x - 5

2x² - 11x + 5 = 11x - 5 - 11x + 5

2x-² - 11x + 5

(2x - 1)(x - 5)

The integer

x - 5 = 0

x - 5 + 5 = 0 + 5

x = 5

The answer is x = 5

5 is a positive integer

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2x - 1 = 0

2x - 1 + 1 = 0 + 1

2x = 1

2x/2 = 1/2

x = 1/2

1/2 is a rational number and is not a integer

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Integer numbers are

. . . . -3, -2, -1, 0, 1, 2, 3. . . .

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Answer 2

The equation is 2x^2 = 11x - 5, where x is the required integer. You need to rearrange the above equation and solve for x.

Good luck.

Answer 3

2x^2 + 5 - 11x

2x^2 -11x + 5 = 0

(x-5)(2x-1) = 0

x = 5, 1/2

Answer 4

[2(x^2)]=11x-5

--> 2x^2-11x+5=0

--->(x-5)(2x-1)

---->x=5 or x=1/2 (1/2 is not an interger, so answer must be 5)