Question

For what value of k are the roots of the quadratic equation kx (x-2)+6=0 equal?​

Answer 1

Answer:

$\boxed{\sf k=6}$

Step-by-step explanation:

$\sf kx (x-2)+6=0$

Expand brackets.

$\sf kx^2 -2kx+6=0$

This is in quadratic form.

$\sf ax^2 +bx+c=0$

Since this is for equal roots:

$\sf b^2 -4ac=0$

$\sf a=k\\b=-2k\\c=6$

$\sf (-2k)^2 -4(k)(6)=0$

$\sf 4k^2-24k=0$

$\sf 4k(k-6)=0$

$\sf 4k=0\\k=0$

$\sf k-6=0\\k=6$

Plug k as 0 to check.

$\sf \sf 0x^2 -2(0)x+6=0\\6=0$

False.

So that means k must equal 6.