Question

Show that the equation kx ^{2}-4x+k=2kx-3x ^{2}-1 has real and distinct roots for all real values of k.

Answer 1

\[kx^{2}-4x+k=2kx-3x^{2}-1\]

Answer 2

take the value of the \(\Delta_x\). You'll get a value independent from \(k\)

Answer 3

restructure it into a quadratic and observe the determinant of the quadratic formula

Answer 4

a=k+3 b=-4-2x c=k+1 but what to do from here ..?

Answer 5

if b^2 -4ac >0 you have real distinct roots

Answer 6

and thats -4-2k just in case that wasnt a typo

Answer 7

(-4-2k)^2 -4(k+3)(k+1) > 0

(4k^2+16+16k) -4(k^2+4k+3) > 0

4k^2+16+16k -4k^2-16k-12 > 0

16+16k-16k-12 > 0

16-12 > 0

4 > 0 is always true regardless of k