Respuesta :

To figure ut the roots use the quadratic formula

x = [-b +- sqrt(b^2-4ac)]/2a

x = [-k +- sqrt(k^2-4(1)(5)]/2(1)

x = [-k + sqrt(k^2 - 20)]/2 or [-k - sqrt(k^2 - 20)]/2 

So the question says these roots differ by sqrt 61, so let's subtract each

[-k + sqrt(k^2 - 20)]/2 - [-k - sqrt(k^2 - 20)]/2 

well the k's cancel in the beginning and we are left with 2sqrt(k^2 - 20)/2, and the 2 on top and bottom reduce to 

sqrt(k^2 - 20), so this equals sqrt 61

Set equal and solve

sqrt(k^2 - 20) = sqrt(61)

k^2 - 20 = 61

k^2 = 81, so k = +9 or -9

The greatest value therefore is  k = +9.

The possible greatest value is 9.

Given that

The roots of the equation [tex]\rm x^2+kx+5[/tex] differ by [tex]\sqrt{61[/tex].

We have to determine

The greatest possible value of k.

According to the question

The roots of the given equation are determined by the following formula;

[tex]\rm x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

The roots of the equation [tex]\rm x^2+kx+5[/tex] differ by [tex]\sqrt{61[/tex].

Substitute the value in the formula.

[tex]\rm x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\rm \sqrt{61}= \dfrac{-k\pm \sqrt{k^2-4\times 1 \times 5}}{2\times 1}\\\\\rm \dfrac{-k \pm \sqrt{k^2-20}}{2} = \sqrt{61}\\\\{-k \pm \sqrt{k^2-20} = 2\sqrt{61}\\\\[/tex]

Solving the equation for the value of k

[tex]\rm \sqrt{k^2-20} = \sqrt{61}\\\\Squaring \ on \ both \ sides\\\\k^2-20 = 61\\\\k^2 = 61+20\\\\k^2 = 81\\\\k = 9[/tex]

Hence, the possible greatest value is 9.

To know more about the Quadratic equation click the link given below.

https://brainly.com/question/11441437

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