willbeatthegmat wrote:Is underroot of X a prime num?
1)|3x-7| =2x+2
2)x^2= 9x
When we have complex absolute value equations to solve, it's often very useful to square both sides.
(1) |3x-7| = 2x+2
Squaring both sides:
9x^2 - 42x + 49 = 4x^2 + 8x + 4
5x^2 -50x + 45 = 0
5(x^2 -10x +9) = 0
5(x - 9)(x - 1) = 0
x = 9 or x = 1
Is sqrt(x) prime? If x=9 yes, if x=1 no... insufficient. Eliminate (a)/(d).
[for sfd: "underroot x" means the positive square root, i.e. putting x under the root sign]
(2) x^2 = 9x
x^2 - 9x = 0
x(x-9) = 0
x = 0 or x = 9
Is sqrt(x) prime? If x=9 yes, if x=0 no... insufficient. Eliminate (b).
Combined: if both equations are true, x=9. Is sqrt9 prime? YES... choose (c) together sufficient, apart insufficient.
If you chose (b), you probably made a VERY common algebra error. Let's examine (2) again:
x^2 = 9x
Your first inclination might be to "cancel out" an x from both sides, leaving you with:
x = 9
However, you need to remember that when we "cancel out", what we're really doing is dividing both sides by x. What number are we never allowed to divide by? 0! So if you divide both sides by x, you're assuming that x does not equal 0.
If we don't have any limitations on x, then x certainly could be 0. So, if you want to divide through by x, you need to consider 2 cases:
1) x not= 0. If x doesn't equal 0, then we can simplify to x=9.
2) x = 0. If x does equal 0, then... well.. x = 0.
So, we have two possibilities, x=0 and x=9.
In the long run, it's much safer to rearrange the statement just as we would a quadratic:
x(x-9) = 0
which allows us to easily see the two roots.
