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  • Find the value of k for which both roots of the equation <img alt="x^{2}-4 k x-4- k+9 k^{2}=0" src="http://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B2%7D-4%20k%20x-4-%20k&plus;9%20k%5

Find the value of k for which both roots of the equation x^{2}-4 k x-4- k+9 k^{2}=0 are positive

Option: 1

k\in\left (-\frac{4}{5},1 \right )


Option: 2

k\in\phi


Option: 3

k>\frac{5}{2}


Option: 4

k\in(-\infty,\frac{7-\sqrt{135}}{6})\cup(\frac{7+\sqrt{135}}{6},\infty)


Answers (1)

best_answer

Location of roots (1)

If both roots of f(x) are greater than k

 

     



 

i) D ≥ 0

ii) af(k) > 0

iii) \\\mathrm{k < \frac{-b}{2a}  

 

Now,

If both the roots of equation x^{2}-4 k x-4- k+9 k^{2}=0 are positive, it means that both roots of the equation are greater than 0. So k = 0.

  • D \geq 0,
  • af(0)>0 and
  •  0 < \frac{-b}{2a}

i) D \geq 0

\\( -4k)^{2}-4\left(-4- k+9 k^{2}\right)\geq0\\16k^2+16+4k-36k^2\geq0\\ 5k^2-k-4\leq0\\ k\in\left [-\frac{4}{5},1 \right ]

ii) a.f(0)>0

\\0 -4k(0) -4-k+9k^2 > 0

 

From (i), (ii) and (iii)

k\in\phi or for no values of k sum of the roots is greater than 5

 

Posted by

Kuldeep Maurya

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