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Factor original equation to k(k-1) Substitute k with k -1 to get (k-1)(k-1) = 0 k =1, sufficient

2. Solve for k k = (sqrt(5) + 1)/2 No need to solve as it will be a value that you plug into original equation. Sufficient sqrt(5) will be a positive number as it's an even root, so no need to worry about negative value.

Last edited by davidkwu on 19 Dec 2014, 14:43, edited 1 time in total.

here we need the value of k^2-k Statement 1 k-1/k=1 k^2-1=k k^-k=1 hence sufficient Statement 2 here we have k so we can get k^2 so we can get k^2-k hence sufficient

Oh! Can the question be re-written? If I think of St 1: (k-1)/ (whole) over K the answer is insuff If it is k - (1/k) it is suff

Thank you!

You should be familiar with basic math notations.

k - 1/k can ONLY mean \(k - \frac{1}{k}\). If it were \(\frac{k - 1}{k}\), it would have been written as (k - 1)/k.

Hi Bunuel,

Thanks for a great question. I thought I shuld not get take the K to other side unless I know the sign. What if it K was negative. Is the rule of not taking variable applicable only in case of equality signs (<,>). Can I take the K to other side when it is equal to(=).

Oh! Can the question be re-written? If I think of St 1: (k-1)/ (whole) over K the answer is insuff If it is k - (1/k) it is suff

Thank you!

You should be familiar with basic math notations.

k - 1/k can ONLY mean \(k - \frac{1}{k}\). If it were \(\frac{k - 1}{k}\), it would have been written as (k - 1)/k.

Hi Bunuel,

Thanks for a great question. I thought I shuld not get take the K to other side unless I know the sign. What if it K was negative. Is the rule of not taking variable applicable only in case of equality signs (<,>). Can I take the K to other side when it is equal to(=).

We cannot multiply/divide an inequality by the variable if we don't know its sign but we can add/subtract whatever we want to/from both sides of an inequality.

For example, we cannot divide x > y by x unless we know the sign of x. If x is positive, then we'll get 1 > y/x but if x is negative, then we'll get 1 < y/x (flip the sign when multiplying/dividing by negative number). On the other hand we can subtract x from both sides of x - x > y - x to get 0 > y - x.