The set of values of \(x\) for which \[x^2 + bx + c \lt 0\] is \(-2 \lt x \lt 7\), where \(b\) and \(c\) are constants. Find the value of \(b\) and the valu...

Nine marks built on one idea: a quadratic inequality and the roots of the quadratic are two views of the same object. If \(x^2 + bx + c \lt 0\) exactly on \(-2 \lt x \lt 7\), then the endpoints of that interval must be the roots, because a positive quadratic is negative precisely between its roots. That single observation makes part (a) immediate.

(a) The values of \(b\) and \(c\) [3 marks]

The roots are \(-2\) and \(7\), and the coefficient of \(x^2\) is 1, so the quadratic factorises as \[x^2 + bx + c = (x + 2)(x - 7) \quad \textbf{M1}\] Expanding, \[= x^2 - 7x + 2x - 14 = x^2 - 5x - 14 \quad \textbf{A1}\] \[b = -5, \qquad c = -14 \quad \textbf{A1}\] The M1 is for using the interval endpoints as roots. An equally valid route uses sum and product of roots: \(-b = (-2) + 7 = 5\) so \(b = -5\), and \(c = (-2)(7) = -14\). Either way the final A1 needs both constants, with the signs right. The commonest slip is writing \((x - 2)(x + 7)\), which corresponds to the interval \(-7 \lt x \lt 2\) instead.

(b) Solve \(x^2 + bx + c \ge 4x - 14\) [3 marks]

Substitute the constants and collect everything on one side. The \(-14\) appears on both sides and cancels, which is the point of the numbers chosen: \[x^2 - 5x - 14 \ge 4x - 14 \ \Rightarrow\ x^2 - 9x \ge 0 \quad \textbf{M1}\] \[x(x - 9) \ge 0 \quad \textbf{A1}\] The critical values are \(0\) and \(9\), and since the parabola opens upwards it is at or above the axis outside its roots: \[x \le 0 \quad \text{or} \quad x \ge 9 \quad \textbf{A1}\] Note that \(x = 0\) is a genuine solution here, so the inequality is inclusive at both ends. Do not divide through by \(x\); that would lose the root \(x = 0\) and is invalid because \(x\) may be negative or zero.

(c) Solve \(\dfrac{x^2 + bx + c}{x + 1} \le 0\) [3 marks]

Use the factorised numerator from part (a). A rational expression changes sign at a zero of the numerator and at a zero of the denominator, so all three matter: \[\frac{(x + 2)(x - 7)}{x + 1} \le 0, \qquad \text{critical values } x = -2, \ -1, \ 7 \quad \textbf{M1}\] Build a sign table over the four regions. To the left of \(-2\) the two numerator factors are negative and the denominator is negative, giving an overall negative value; each subsequent critical value flips the sign: \[\lt 0 \ \text{for} \ x \lt -2; \quad \gt 0 \ \text{for} \ -2 \lt x \lt -1; \quad \lt 0 \ \text{for} \ -1 \lt x \lt 7; \quad \gt 0 \ \text{for} \ x \gt 7 \quad \textbf{A1}\] The two numerator roots may be included, because the fraction is then exactly zero; the denominator root must be excluded, because the expression is undefined there: \[x \le -2 \quad \text{or} \quad -1 \lt x \le 7 \quad \textbf{A1}\] Notice the asymmetry in the final answer: an inclusive bound at \(-2\) and at \(7\), but a strict one at \(-1\). Getting that right is exactly what the last A1 is testing.

Common wrong turns on this question

• Multiplying through by \((x + 1)\) in part (c). That expression is negative for \(x \lt -1\), so the inequality would have to reverse there; the safe move is a sign table, or multiplying by \((x + 1)^2\).
• Ignoring the denominator when finding critical values and reporting \(-2 \le x \le 7\).
• Including \(x = -1\) in the solution set.
• Dividing by \(x\) in part (b) and losing the region \(x \le 0\).
• Getting the sign of \(b\) wrong in part (a) by confusing the roots with the coefficients.

How to check

Part (a) is self-checking: \((-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 0\) and \(7^2 - 35 - 14 = 0\), so both endpoints are roots as required. For part (c), test one value per region: at \(x = -3\) the value is \(\dfrac{(-1)(-10)}{-2} = -5 \le 0\), so that region is in; at \(x = -1.5\) it is \(\dfrac{(0.5)(-8.5)}{-0.5} = 8.5\), so that region is out; at \(x = 0\) it is \(-14 \le 0\), in; at \(x = 8\) it is \(\dfrac{(10)(1)}{9} \gt 0\), out. That matches the answer exactly.