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

Question

anonymous · Answer

$$kx^{2}-4x+k=2kx-3x^{2}-1$$

badhi · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

(-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