Prove that 3x2 = 2k-(2k-3)x has real roots for all real values of k.
let f(x) = 3x2+(2k-3)x-2k
Always arrange equation into ax2+bx+c format
= (2k-3)2 – 4(3)(-2k)
- Always factorise and look for the roots when you see 4k2+9-12k+24k and draw out graph.
- ***Note that b2-4ac curve is not the same as f(x) curve
Since f(x) discriminant b2-4ac ≥ 0 for all values of k, 3x2 = 2k-(2k-3)x has real roots for all values of k.