If the value of -2k+ (4-15k)^1/2 is positive, which of the following

Question

If the value of - 2k+ \sqrt{4-15k} is positive, which of the following ranges represents the value of k ? A. −4k or k>\frac{1}{4} E. −4

imeanup · Answer

-2k + \sqrt{4 - 15k} = 0

2k = \sqrt{4 - 15k}

(2k)^2 = (\sqrt{4 - 15k})^2

4k^2 = 4 - 15k

4k^2 + 15k - 4  = 0

4k^2 + 16k - k - 4  = 0

4k(k + 4) - 1(k + 4)  = 0

(4k - 1)(k + 4)  = 0

k  = -4, \dfrac{1}{4}

A

learnwithrohan · Answer

−2k+ \sqrt{4−15k}

by solving , we do get 
-4 < k < 1/4

however , by looking at the question we can say that : 
for any negative value of K the answer will be positive.

i.e  -2k ( will be positive for all values when k is negative ) and \sqrt{4−15k} will give positive answer anyways 
hence for k<0 the answer should be positive.

so final answer should be 
after combining the two i.e   k<0 and -4 < k < 1/4

k < 1/4

Edoua · Answer

Answer D
4k^2 +15k -4 >0
(K+4)(K-1/4)>0

-4 and 1/4 are the 2 solutions of the equation (k+4)(k-1/4)=0

Using the waiving line method to find the solution

Posted from my mobile device

sayan640 · Answer

KarishmaB   MartyMurray  , After solving -2*k + root over ( 4 - 15*k ) I am getting the range as -4 <k <1/4 and selected option A . What mistake did I make ? Can you please help ?
I found ( k+4 ) * (4k -1 ) < 0 and then used the wavy line method to get suitable range. Please help.

Regor60 · Answer

Try 0.

Equation then becomes:

√4 = 2.

This eliminates B, D and E since 0 is not in their ranges.

Leaving A and C.

How do their ranges differ ?

C allows for all values less than 1/4, while A limits C's range to >-4.

So if -4 generates a positive result, we then could eliminate A:

(-2*-4) + √(4-(15*-4)) =

8 + √64 = 16

Answer C

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MartyMurray · Answer

Notice that  when you start off, you do not actually have this:

-2k + \sqrt{4 - 15k} = 0

What you actually have is this:

-2k + \sqrt{4 - 15k} > 0

Then, this:

\sqrt{4 - 15k} > 2k

So, when you take this step, you are squaring an inequality, which may not be a valid operation if the two sides can have different signs:

(\sqrt{4 - 15k})^2 > (2k)^2

In fact, we can confirm that something is wrong with the approach of squaring both sides and using the wavy line method by noticing that, for any negative k with a large absolute value,   -2k + \sqrt{4 - 15k}  must clearly be positive.

So, you could either use a different approach, such as plugging in values, or use the approach you used and then check the values you get to adjust your result as needed to adjust for whatever might go wrong as a result of squaring the inequality.

I personally find that the easiest way to solve the problem is to see that any negative k with a large absolute value must work. So, A, B, and E are out because they restrict k to above -4 or above 0. Then, notice that any large positive k won't work. So, D is out. Thus, the only possible correct answer is C.

Oppenheimer1945 · Answer

If you're familiar with plotting parabolas, you can solve it within seconds.

f(x)=rt(4-15x)  will be leftwards opening

intersects with  g(x)=2x  at x=1/4.

The function  f(x)  is above  g(x)  for all values of  x<1/4.

Using desmos for clarity.

K-ja · Answer

I am still confused why is the answer C and not A?

sayan640 · Answer

Follow the explanation provided by  Regor60 . You ll easily understand.

Posted from my mobile device

Kinshook · Answer

Asked: If the value of - 2k+ \sqrt{4-15k} is positive, which of the following ranges represents the value of k ? 4 -15k >=0 k <= 4/15 -2k+ \sqrt{4-15k} > 0 -\sqrt{4-15k} > 2k Case 1: k>=0 4 - 15k > 4kˆ2 4kˆ2 + 15k - 4 < 0 (4k-1)(k+4) <0 -4 < k < 1/4 Case 2: k<0 -2k > 0 \sqrt{4-15k} >=0 -2k+ \sqrt{4-15k} > 0 for all values of k<0 k<0 Combining the above 2 results k<1/4 IMO C

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If the value of -2k+ (4-15k)^1/2 is positive, which of the following

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If the value of -2k+ (4-15k)^1/2 is positive, which of the following

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