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Quadratic Challenging Questions 1/3

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

b2-4ac

= (2k-3)2 – 4(3)(-2k)

= 4k2+9-12k+24k

= 4k2+9+12k

= (2k+3)(2k+3)

  • 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.

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