Find the range of values of pp for which the curve y=px2+2(p+2)x+p+7y = px^2 + 2(p + 2)x + p + 7 has no xx-intercepts.

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Worked solution

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1. State the condition for the curve to have no $x$-intercepts

For the curve to have no xx-intercepts, the equation px2+2(p+2)x+p+7=0px^2 + 2(p + 2)x + p + 7 = 0 must have no real roots. Therefore, the discriminant must be less than zero:
b2−4ac<0b^2 - 4ac < 0

2. Substitute the coefficients into the discriminant formula

[2(p+2)]2−4(p)(p+7)<0[2(p + 2)]^2 - 4(p)(p + 7) < 0

3. Expand and simplify the inequality

4(p2+4p+4)−4p2−28p<04(p^2 + 4p + 4) - 4p^2 - 28p < 0
4p2+16p+16−4p2−28p<04p^2 + 16p + 16 - 4p^2 - 28p < 0
−12p+16<0-12p + 16 < 0

4. Solve the linear inequality for $p$

16<12p16 < 12p
p>1612p > \frac{16}{12}
p>43p > \frac{4}{3}

Answer

p>43p > \frac{4}{3}

AI analysis — may contain mistakes.

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