Show that the quadratic equation 2k (x+1) = 3 - 3x² has real roots for all real values of k.
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Answer:
2k (x+1) = 3 - 3x²
2kx + 2k = 3 - 3x²
2kx + 2k - 3 + 3x² = 0
3x² + 2kx + (2k - 3) = 0
To show that an equation has only real roots, we must show that the discriminant ≥ 0
i.e. b² - 4ac ≥ 0
b² - 4ac = (2k)² - 4(3)(2k - 3)
= 4k² - 24k + 36
= 4(k² - 6k + 9)
= 4(k - 3)²
Since k is real, (k - 3)² ≥ 0
It follows that b² - 4ac ≥ 0
Thus, the equation has real roots for all real values of k (shown)
