Quadratics

Write $ax^2 + bx + c$ as $a(x + p)^2 + q$. Take the coefficient of $x^2$ out of the first two terms, halve the coefficient of $x$ inside, and correct for the square you introduced:

The form answers four P1 questions at once. The vertex is $(-p,\ q)$ — here $(3, -13)$; the line of symmetry is $x = -p$; the minimum value (or maximum, if $a < 0$) is $q$; and the range of the function is $f(x) \ge q$. It also gives the inverse of a restricted quadratic function directly, since $y = a(x+p)^2 + q$ rearranges without the formula.

For $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$ — the part of the quadratic formula under the root — and its sign says how many real roots exist:

“Real roots” without “distinct” means $b^2 - 4ac \ge 0$; the equals sign is the difference between a correct answer and a lost mark. And note what the discriminant is computed on: the quadratic you get after rearranging to “= 0”, with its own $a$, $b$, $c$ — which in a “find $k$” question will contain $k$.

To solve $ax^2 + bx + c > 0$: find the critical values where the expression is zero, sketch the parabola (up if $a > 0$), and read off where it is above or below the axis. For $a > 0$, the expression is negative between the roots and positive outside them:

The sketch is not optional working — it is how the direction of the inequality is decided, and examiners accept it as the justification. The outside case is two separate regions joined by “or”; writing it as $3 < x < 2$ is a common and costly slip.

The P1 question type. Substitute the line into the curve, rearrange to a quadratic in $x$ equal to zero, then apply the discriminant condition the wording implies. The coefficient of $x$ — and sometimes the constant — will contain the unknown $k$, so the discriminant becomes a quadratic inequality in $k$, solved as above.

P1 hides quadratics inside other functions and expects a substitution:

• $x^4 - 5x^2 + 4 = 0$: let $u = x^2$, so $u^2 - 5u + 4 = 0$, $u = 1$ or $4$, hence $x = \pm 1, \pm 2$.
• $x - 3\sqrt{x} + 2 = 0$: let $u = \sqrt{x}$, so $u^2 - 3u + 2 = 0$, $u = 1$ or $2$, hence $x = 1$ or $4$ — and $u \ge 0$, so a negative $u$ would be rejected.
• $\tan^2\theta - 3\tan\theta + 2 = 0$: let $u = \tan\theta$; the quadratic gives the values of $\tan\theta$, and the trigonometry finds $\theta$.

The mark that is lost here is the last one: solving for $u$ and stopping, or forgetting that $x^2 = 4$ has two solutions.

1. 1.Completing the square with a ≠ 1 without factorising first

   Take $a$ out of the $x^2$ and $x$ terms before halving. The correction term is then $a \times (\tfrac{b}{2a})^2$, and most sign errors come from skipping that step.

2. 2.Strict versus non-strict inequality

   “Two distinct roots” is $>$; “real roots” is $\ge$; “tangent” is $=$. Match the sign to the wording before solving.

3. 3.Discriminant of the wrong quadratic

   It applies to the equation after substitution and rearrangement to zero — not to the original curve. Identify $a, b, c$ from that equation.

4. 4.Inequality solved without a sketch

   Critical values alone do not tell you which region. Sketch the parabola; between the roots for negative, outside for positive (when $a > 0$).

5. 5.Stopping at the substitution variable

   $u = 4$ is not the answer to an equation in $x$. Substitute back, and take both square roots where they exist.