Question

plz do this it is a quadratic equation. Answer will come in terms of k. I will mark as brainliest if the answer is right

Answer 1

Given Equation is 6x^2 = 11kx + 7k^2

= > 6x^2 - 7k^2 = 11kx

= > 6x^2 - 11kx - 7k^2  = 0

It is in the form of ax^2 + bx + c = 0,

Where a = 6, b = -11k, c = -7k^2.

The solutions are:

(1)

$x = \frac{-b+ \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ x = \frac{-(11k)+ \sqrt{(-11k)^2 - 4 * 6 * (-7k)^2} }{2 *6}$

$= \ \textgreater \ \frac{11k + \sqrt{11k^2 + 168k^2} }{12}$

$= \ \textgreater \ \frac{11k + \sqrt{121k^2 + 168k^2} }{12}$

$= \ \textgreater \ \frac{11k + \sqrt{289k^2} }{12}$

$= \ \textgreater \ \frac{11k + 17k}{12}$

$= \ \textgreater \ \frac{28k}{12}$

$= \ \textgreater \ x = \frac{7k}{3}$



(2)


$x = \frac{-b- \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(-11k) - \sqrt{(-11k)^2 - 4 * 6 * (-7k)^2} }{2 *6}$

$= \ \textgreater \ \frac{11k - \sqrt{121k^2 + 168k^2} }{12}$

$= \ \textgreater \ \frac{11k - \sqrt{289k^2} }{12}$

$= \ \textgreater \ \frac{11k - 17k}{12}$

$= \ \textgreater \ \frac{-6k}{12}$

$= \ \textgreater \ x = \frac{-k}{2}$



Therefore x = (7k)/3, x = -k/2.



Hope this helps!

Answer 2

Hi,

Please see the attached file!


Thanks