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If 'x' is more than 1, then x2 is more than 'x'.

If 'x' is less than 1, then x2 is less than 'x'.

Q: Is x2 less than x

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x2 - 15 = 2x x2 - 2x - 15 = 0 x2 - 2x - 15 = (x - 5)(x + 3) x = 5, -3

if you are doing algebra you have to have a number on the other side of the equal sign so x2-6 has to have a equal sign to figure out 6 is less than x cubed if x is greater than the cube root of 6. The answer is "true" or "false" depending on x

it may equals (n-2)(n+2)

The answer would be 3x if 'X2' is '2x', but if 'X2' is x2, then the total answer would be x2 + x.

Find the value of X in the problem 2X = 16. Solving for X will give you _______.But you don't really want the value of X multiplied by 2 giving you 16. Instead you want the value that's less than or greater than 16. So the answer would really be ± your value of X from above.

Related questions

If x2 < 25 Then: |x| < 5 -5 < x < 5

(x - 2)(x + 3) = x2 + 3x - 2x -6 = x2 + x - 6

That factors to (x - 6)(x + 3) x is less than or equal to -3, 6

x2-2x = 15 x2-2x-15 = 0 (x-5)(x+3) = 0 Therefore: x = 5 or x = -3

When x is a number between -1 and 0. For example, if x=-0.5: x < x3 < x2 because -0.5 < -0.125 < 0.25

x2 - 15 = 2x x2 - 2x - 15 = 0 x2 - 2x - 15 = (x - 5)(x + 3) x = 5, -3

It depends on whether the number being raised to the power is greater than or less than 1. If x < 1 then x3 < x2 If x > 1 then x3 > x2 with equality at x = 1

You don't. There's no question there, and nothing that needs solving. "16 less than x squared" is written as (x2 - 16)

No because its discriminant is less than zero

Call the numbers: x, (x+1), (x+2) Product of the smaller two: x(x+1) = x2 + x Square of the largest number: (x+2)2 = x2 + 4x + 4 Product of the smaller two is 40 less than the largest one squared: (x2 + x) = (x2 + 4x + 4) - 40 x2 + x = x2 + 4x - 36 0 = 3x - 36 0 = x - 12 x = 12 x+1 = 13 x+2 = 14 Check: Product of the smaller two = (12)(13) = 156 Square of the largest = 196. 156 is in fact 40 less than 196.

It can't be factored because its discriminant is less than zero

There are no solutions because the discriminant is less than zero