Question

Given that y = 4kx^2 - 6x + (k-4), find the values of k, for which the equation y = 0 has real roots.

So i did b^2 - 4ac>0 and i got (1 + 2k)(9 - 2k)>0

Now what??

Answer 1

you have factorised correctly so look at k ignoring the inequality

the solutions are k = -1/2 and 9/2

I would pick an value between the k values to test

say k = 0

then the quadratic (1 + 2k)(9 - 2k)> 0 when k = 0

so the inequality is

\[-\frac{1}{2} < k < \frac{9}{2}\]

Answer 2

YUP i got that :)