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Worked solution

Find when a quadratic is strictly positive everywhere

Find the range of values of kk for which 3x2+6x+k3x^2+6x+k is always positive for all real values of xx.

CJC · O Level revision · Example 4

Answer

k>3k>3.

Compare your reasoning

Check the leading coefficient

b2−4ac<0b^2-4ac<0

The coefficient of x² is 3>0, so the parabola opens upwards. To stay strictly above the axis it must have no real roots.

Solve the strict discriminant inequality

62−4(3)k<0 ⟹ 36−12k<0 ⟹ 12k>36 ⟹ k>36^2-4(3)k<0\ \Longrightarrow\ 36-12k<0\ \Longrightarrow\ 12k>36\ \Longrightarrow\ k>3

Keep the endpoint excluded; equality would make the minimum zero.

How did this question go?

Common mistakes

At k=3 the expression is 3(x+1)² and vanishes at x=−1. Nonnegative is weaker than positive.