the smallest k...

Algebra Level 3

What is the smallest value of k k for which both the roots of the equation x 2 8 k x + 16 ( k 2 k + 1 ) = 0 x^2-8kx+16(k^2-k+1)=0 are real, distinct, and have value at least 4?


The answer is 2.

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3 solutions

Chew-Seong Cheong
Dec 23, 2014

The answer is correct and it can be shown graphically. In fact, I solved it using graphs before I thought out the solution. This was easily done using a Microsoft Excel spreadsheet. I have plotted the curves for k = 0 k=0 , k = 1 k=1 , k = 1.5 k=1.5 , k = 2 k=2 and k = 3 k=3 as in the above to show it. It can be seen that the curve shifts to the right as k k increases. When k = 0 k=0 , there is no real root. When k = 1 k=1 there is only one root as the curve just touches the x x -axis. I also mentioned this in my solution. Of course written equation and graph show the same thing. When 1 < k < 2 1 < k < 2 , there are two roots but one of the root is less than 4 4 , when x = 2 x = 2 , the smaller root is exactly 4 4 , as mentioned in my solution. Therefore, when k > 2 k>2 both roots are larger than 4 4 , as shown in the graph with k = 3 k=3 . QED.


The roots of f ( x ) = x 2 8 k x + 16 ( k 2 k + 1 ) f(x) = x^2-8kx+16(k^2-k+1) are:

x = 8 k ± 64 64 ( k 2 k + 1 ) 2 = 4 k ± 4 k 1 x = \dfrac {8k \pm \sqrt{64-64(k^2-k+1)}} {2} = 4k \pm 4\sqrt{k-1}

For the roots to be at least 4 4 , then:

4 k 4 k 1 4 k k 1 1 k 1 k 1 4k - 4\sqrt{k-1} \ge 4\quad \Rightarrow k - \sqrt{k-1} \ge 1 \quad \Rightarrow k - 1 \ge \sqrt{k-1}

k 2 2 k + 1 k 1 k 2 3 k + 2 0 ( k 1 ) ( k 2 ) 0 \Rightarrow k^2 -2k + 1 \ge k-1 \quad \Rightarrow k^2 -3k + 2 \ge 0 \quad \Rightarrow (k-1)(k-2) \ge 0

k 1 \Rightarrow k \ge 1\space or k 2 \space k \ge 2 .

We find that when k = 1 k=1 there is only one root 4 ± 0 4 \pm 0 , therefore it is not acceptable.

When k = 2 k=2 , the roots are { 4 , 12 } \{4,12\} , therefore the answer is k 2 k \ge \boxed {2} .

Elegant solution,sir!

Priyatam Roy - 6 years, 5 months ago

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Actually, excellent question.

Chew-Seong Cheong - 6 years, 5 months ago

nice solution,sir. is the graph necessary ??

Aareyan Manzoor - 6 years, 5 months ago

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I used the graphs to explain to Priyatam Roy.

Chew-Seong Cheong - 6 years, 5 months ago

Really sir nice solution !!!!!!!!!!!!

Abhisek Mohanty - 6 years, 2 months ago
Mehul Chaturvedi
Dec 28, 2014

The roots of this equation can be written as

8 k ± 64 k 2 64 k 2 + 64 k 64 2 8 k ± 8 k 1 2 4 k ± 4 k 1 \dfrac { 8k\pm \sqrt { 64{ k }^{ 2 }-64{ k }^{ 2 }+64k-64 } }{ 2 } \\ \Rightarrow \dfrac { 8k\pm 8\sqrt { k-1 } }{ 2 } \\ \Rightarrow 4k\pm 4\sqrt { k-1 }

Now let us put values

  • k k can't be 1 1 as it would result equal roots

  • so let k k = 2 2

Yes it gives us 4 × 2 ± 4 2 1 8 4 4 \Rightarrow 4\times 2\pm 4\sqrt { 2-1 } \\ \Rightarrow ~ 8-4\\ \Rightarrow \color{#69047E}{4}

So the answer is 1 \color{#20A900}{\huge\boxed{1}}

Aritra Jana
Dec 23, 2014

consider the equation graphically: (This picture does not depict the exact same equation given in the question)

Because the quadratic is monic, it will represent a parabola facing upwards with vertex downwards. It has been mentioned that the roots are both greater than 4.

therefore, the value of the function that represents the quadratic will be greater than 0 at x≤4 \text{the value of the function that represents the quadratic will be greater than 0 at x≤4} this is actually very simple: for the typical quadratic y = x 2 + b x + c y=x^{2}+bx+c ,for any value of x x that lies in the interval containing the roots, the curve will be below the X-axis- thus ≤0 and the curve will be above the X-axis anywhere else.

So, plugging in x = 4 x=4 we should get the value of the function≥0

f ( 4 ) = 16 32 k + 16 ( k 2 k + 1 ) 0 k 2 3 k + 2 0 k ( , 1 ] [ 2 , + ) \therefore f(4)=16-32k+16(k^{2}-k+1)≥0 \implies k^{2}-3k+2≥0 \implies k\in(-\infty,1] \cup [2,+\infty)

it is also mentioned that the roots are real and distinct, thus, the discriminant Δ > 0 \Delta>0

checking: Δ = 64 k 2 64 ( k 2 k + 1 ) > 0 k > 1 \Delta=64k^{2}-64(k^{2}-k+1)>0 \implies k>1

therefore from the two results, we get k 2 \large{k≥\boxed{2}}

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