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Quadratic functions and their graphs, completing the square, and the discriminant \(b^2 - 4ac\), which tells you how many real roots a quadratic has. You also solve quadratics in a function of \(x\), such as \(x^4 - 10x^2 + 9 = 0\). Discriminant questions with an unknown constant \(k\) are very common and worth 4 to 6 marks.
Key facts
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Discriminant: \(b^2 - 4ac\)
\(\gt 0\): two distinct real roots; \(= 0\): repeated root; \(\lt 0\): no real roots; \(\ge 0\): real roots
The roots of \(ax^2 + bx + c = 0\) are where the graph of \(y = ax^2 + bx + c\) meets the \(x\)-axis.
Solve by factorising, by the formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), or by completing the square.
\(a \gt 0\): a \(\cup\)-shaped curve with a minimum point. \(a \lt 0\): a \(\cap\)-shaped curve with a maximum point. The \(y\)-intercept is \((0, c)\)
\(y = a(x + p)^2 + q\) has turning point \((-p, q)\) and line of symmetry \(x = -p\)
In completed-square form you can read off the turning point (3, −13) and the line of symmetry x = 3.
If \(a \gt 0\), the minimum value is \(q\), so if \(q \gt 0\) the expression is always positive.
The discriminant
diagram
The discriminant of \(ax^2 + bx + c\) is \(b^2 - 4ac\)
\(b^2 - 4ac \gt 0\): two distinct real roots (the curve crosses the \(x\)-axis twice).
\(b^2 - 4ac = 0\): one repeated root (the curve touches the \(x\)-axis).
\(b^2 - 4ac \lt 0\): no real roots (the curve does not meet the \(x\)-axis).
The discriminant tells you how many times the curve meets the x-axis.
'Real roots' includes the repeated case, so it means \(b^2 - 4ac \ge 0\)
With an unknown \(k\): write down \(a\), \(b\) and \(c\) in terms of \(k\), form the equation or inequality, and solve it.
Quadratics in a function of x
\(x^4 - 10x^2 + 9 = 0\) is \(u^2 - 10u + 9 = 0\) with \(u = x^2\), so \(u = 1\) or \(9\), giving \(x = \pm 1\) or \(\pm 3\)
\(x - 5\sqrt{x} + 6 = 0\) uses \(u = \sqrt{x}\): \(u = 2\) or \(3\), so \(x = 4\) or \(9\)
Reject impossible values: \(x^2 = -2\), \(\sqrt{x} = -1\) and \(2^x = -3\) have no real solutions.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, ft = follow through: you still earn the mark if you use your own earlier wrong answer correctly.
Complete the square and find the turning point
3 to 4 marksGrade D
Take out the coefficient of \(x^2\) from the \(x\) terms.
Halve the coefficient of \(x\) inside the bracket and subtract its square (multiplied by \(a\)).
Read the turning point from \(a(x + p)^2 + q\): it is \((-p, q)\)
Example. (a) Write \(3x^2 + 12x + 5\) in the form \(a(x + b)^2 + c\). (b) Hence state the coordinates of the minimum point of the curve \(y = 3x^2 + 12x + 5\)
Show the model answerHide the model answer
(a) \(3(x^2 + 4x) + 5\), so \(a = 3\)
(B1)
\(3(x + 2)^2 - 12 + 5\)
(M1)
\(3(x + 2)^2 - 7\)
(A1)
(b) \((-2, -7)\)
(B1 ft)
Find the values of k using the discriminant
4 marksGrade B
Identify \(a\), \(b\) and \(c\) in terms of \(k\), using brackets.
Use the right condition: \(\gt 0\),\(= 0\), \(\ge 0\) or \(\lt 0\)
Simplify to the printed inequality if there is one.
Solve the quadratic inequality in \(k\) using a sketch.
Example. The equation \(x^2 + (k - 3)x + (3 - 2k) = 0\), where \(k\) is a constant, has two distinct real roots. (a) Show that \(k^2 + 2k - 3 \gt 0\). (b) Find the set of possible values of \(k\)
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(a) Two distinct real roots, so \(b^2 - 4ac \gt 0\): \((k - 3)^2 - 4(1)(3 - 2k) \gt 0\)
(c)Answer: \(k^2 - 2k + 9 = (k - 1)^2 + 8 \ge 8 \gt 0\), so there are always two distinct real roots.
M1 for completing the square: \(k^2 - 2k + 9 = (k - 1)^2 + 8\)
A1 for \((k - 1)^2 \ge 0\) so the discriminant is at least 8, hence positive
A1* for a conclusion: discriminant \(\gt 0\) for all \(k\), so two distinct real roots and two distinct points of intersection
Worked solution: \(k^2 - 2k + 9 = (k - 1)^2 + 8\). Since \((k - 1)^2 \ge 0\), the discriminant is at least 8, so it is always positive. So the quadratic always has two distinct real roots, and \(l\) meets \(C\) at two distinct points for every real \(k\).
Question 3Hard9 marks
The curve \(C\) has equation \[y = \frac{x^2 + x + 1}{x^2 + 1}, \quad x \in \mathbb{R}\]
(a) Show that, for a given value of \(y\), the \(x\)-coordinates of the points on \(C\) with that \(y\)-coordinate satisfy \[(y - 1)x^2 - x + (y - 1) = 0\][2]
(b) Hence prove that \(\frac{1}{2} \le y \le \frac{3}{2}\) for every point on \(C\).[5]
(c) Find the coordinates of the points on \(C\) where \(y\) takes its greatest and least values.[2]
M1 for multiplying both sides by \(x^2 + 1\) (which is never zero)
A1* for collecting terms to obtain the given equation
Worked solution: \(x^2 + 1 \gt 0\), so \(y(x^2 + 1) = x^2 + x + 1\), i.e. \(yx^2 + y - x^2 - x - 1 = 0\), which is \((y - 1)x^2 - x + (y - 1) = 0\).
(b)Answer: For \(y \ne 1\), a real \(x\) needs \(1 - 4(y - 1)^2 \ge 0\), so \(|y - 1| \le \frac{1}{2}\); \(y = 1\) (at \(x = 0\)) is also in the interval.
M1 for using the condition for real roots: \((-1)^2 - 4(y - 1)(y - 1) \ge 0\)
M1 for solving: \(-\frac{1}{2} \le y - 1 \le \frac{1}{2}\) (or critical values \(\frac{1}{2}\), \(\frac{3}{2}\) with the inside region)
A1 for \(\frac{1}{2} \le y \le \frac{3}{2}\)
B1 for dealing with \(y = 1\), where the equation is linear (\(x = 0\)), and noting it lies in the interval; with a conclusion
Worked solution: Any point on \(C\) has a real \(x\)-coordinate, which is a root of \((y - 1)x^2 - x + (y - 1) = 0\). If \(y \ne 1\) this is a quadratic, so its discriminant is non-negative: \(1 - 4(y - 1)^2 \ge 0\), so \((y - 1)^2 \le \frac{1}{4}\) and \(-\frac{1}{2} \le y - 1 \le \frac{1}{2}\), i.e. \(\frac{1}{2} \le y \le \frac{3}{2}\). If \(y = 1\) the equation is \(-x = 0\), so \(x = 0\); and \(y = 1\) lies in the interval. So \(\frac{1}{2} \le y \le \frac{3}{2}\) for every point on \(C\).
(c)Answer: \(\left(1, \frac{3}{2}\right)\) and \(\left(-1, \frac{1}{2}\right)\)
M1 for substituting \(y = \frac{3}{2}\) or \(y = \frac{1}{2}\) into the equation in (a) and solving (a repeated root)
A1 for \(\left(1, \frac{3}{2}\right)\) and \(\left(-1, \frac{1}{2}\right)\)
Worked solution: \(y = \frac{3}{2}\): \(\frac{1}{2}x^2 - x + \frac{1}{2} = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1\). \(y = \frac{1}{2}\): \(-\frac{1}{2}x^2 - x - \frac{1}{2} = 0 \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1\). Greatest value at \(\left(1, \frac{3}{2}\right)\), least value at \(\left(-1, \frac{1}{2}\right)\).