Inequality 9

Is underroot of X a prime num?
1)|3x-7| =2x+2
2)x^2= 9x

From 1: If x>0 then 3x-7=2x+2 or x=9;
If x<0 then -3x+7=2x+2 or x=1 (But x<0)
Hence only value of x=9
Sufficient

From 2: x^2=9x or x^2-9x=0 or x(x-9)=0;

Thus x=0;or 9 Here we get 2 values of x and can't conclude about the value of x.Thus Insufficient

Hence Ans=A

Vikram there is flaw in your argument

If x<0 then -3x+7=2x+2 or x=1 (But x<0)

Why did you make equation -3x-7 to -3x+7...?

Since only the value of x is changing you cannot change the sign of 7...

So the equation is -3x-7=2x+2

-5x=9 or x=-9/5

Hence its C

yalanand wrote:Vikram there is flaw in your argument

If x<0 then -3x+7=2x+2 or x=1 (But x<0)

Why did you make equation -3x-7 to -3x+7...?

Since only the value of x is changing you cannot change the sign of 7...

So the equation is -3x-7=2x+2

-5x=9 or x=-9/5

Hence its C

Point Taken.Thanks.

Can one of you please tell me what an underroot is?

I worked slightly in a slightly different way to Vikram and yalanand, but I could be wrong:


|3x-7| = 2x+2

When (3x-7) is +ve:

3x-7 = 2x+2
3x-9 = 2x
x-9 = 0
x = 9, and sq. rt. x is prime

When (3x-7) is -ve:
-(3x-7) = 2x+2
-3x+7 = 2x+2
-3x+5=2x
5=5x
x=1, and sq. rt. x is *not* prime

therefore 1) is not sufficient

2) x^2 = 9x implies x=0 or x=9, not sufficient

But 1) and 2) together allow x to have only one value, which is sufficient to answer the question. So it's C.

What I want to know is, is my method of considering (1) accurate, where I can consider the positive and negative possibilities for (3x-7), rather than just changing x?

answer is C...watever has been stated by cjb is correct but 1 thing.

wen u take a negative value in a mod first we need to consider the valid values of x.

here for neg val. of |3x-7| we need to find valid values for x...

3x-7 < 0....coz we ant this to be negative.
hence x<7/3 are the set of valid values for x for |3x-7| to be negative.

so wen we take negative value -3x +7 =2x + 2; x= 1....
now x<7/3...so x=1 satisfies...had the value came out to be x>7/3.....the answer wud have been A....so it wud hav made a diff to the answer..
just that in this question the context was right.

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.

Oh, I get it - this is the just the "back check" (I'm sure there's a better phrase for it) re-arranged. It makes sure that the assumption about the sign of the "bracketed" terms is correct. If the term doesn't satisfy the implied inequality (e.g. 3x - 7 > 0) then there's no solution for that sign.

excellent explanation stuart n every1...