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x>3?

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x>3?

by El Cucu » Wed Mar 18, 2009 6:49 am
If x is positive is x>3?

1) (x-1)^2 > 4


2) (x-2)^2 > 9


Hi guys:
My doubt is how to find the roots of the equation when there is an inequality.
I pick number 3 to check if x>3 but I don't know if this is the correct approach.
Many tks in advance.
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by DanaJ » Wed Mar 18, 2009 7:45 am
I think you should just solve it algebraically, it really isn't that complicated.

1. (x - 1)^2 > 4 - take 4 and shift if to the other side to get:
(x - 1)^2 - 4 > 0
Notice that 4 = 2^2 and this is why we have a difference of squares and use the formula a^2 - b^2 = (a - b)(a + b). In this case, a = x - 1 and b = 2. You get that:
(x - 4)(x +1) > 0. This inequality stands only when x is greater than 4 or smaller than -1. The interval will be (-infinite, -1) U (4, infinite).
Remember that x is positive: this eliminates the negative interval above, with x being greater than 4 out final solution. If x > 4, then x will certainly be greater than 3. So 1 is sufficient.

2. Use the same principle (9 = 3^2) to get to:
(x - 5)(x + 1) > 0.
Again, x can be either greater than 5 or smaller than -1. Since x is restricted to positives, then x must be greater than 5 and therefore greater than 3. So 2 is sufficient as well.

Pick D
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