How many integer values of n satisfy the inequality 3√n √n^3 ?

Question

How many integer values of n satisfy the inequality  3√n >√n^3 ?

A. 0

B. 1

C. 2

D. 4

E. Infinitely many

chetan2u · Answer

Now, √n^3 can be written as \sqrt{n*n*n} or n\sqrt{n} 3√n >√n^3 3√n >n√n 3√n -n√n>0 √n(3-n)>0 As \sqrt{n} cannot be negative and n is not 0 from the given inequality, 3-n>0.........n<3 So n is 1 and 2 => two values. C OR 3√n >√n^3 tells us that n cannot be 0, otherwise both sides iwll be 0, and we will have 0>0. Also, we deal with only real numbers, so \sqrt{n} means n has to be non negative. Since n is positive we can cancel out \sqrt{n} from each side. 3√n >√n^3 => 3>√n^2..........3>n .......0

Shrey08 · Answer

3√n >√n^3 √n \geq{0} ....... but since n=0 doesn't satisfy above expression, therefore n > 0 ........... (1) n is always positive...Dividing left hand side and right hand side by √n ..... 3>n ........... (2) From (1) and (2) we know 0

ansarose · Answer

shouldn't it be d) 4 because -2,-1,1 and 2 satisfy the equation?

Bunuel · Answer

-2 and -1 do not satisfy 3√n > √n^3. Also, n cannot be negative because all even roots from negative numbers are undefined on the GMAT.

nsdl1985 · Answer

Wonderful explanation. Thx for the same.

btsaami · Answer

--> Squaring both sides, we get 9n > n^3 n^3 - 9n<0 n(n-3)(n+3) < 0 Place -3, 0 and 3 on the number line. We see that the inequality will be -ve when x<-3 and 0

Mihir_01999 · Answer

Can we take Square on both sides in an inequality ?

Bunuel · Answer

You can square an inequality only when both sides are nonnegative. However, be careful: squaring can introduce extraneous solutions, so when squaring, we should at the same time define the domain and later check the solution against it. Here we have 3√n > √(n^3). We can square because both sides are square roots, which are nonnegative. The domain is n > 0 because even roots are defined for only nonnegative values on the GMAT. Squaring gives 9n > n^3, which gives 0 < n < 3 or n < -3. But since the domain is n > 0, we check the solution against it and are only left with 0 < n < 3.

How many integer values of n satisfy the inequality 3√n > √n^3 ?

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