Quadratics

📚The mathematics

Completing the square

Any quadratic can be written in the form

and this form is worth far more than the expanded one, because it hands you three things at once: the vertex is at \( (-b,\ c) \), the line of symmetry is \( x = -b \), and the least (or greatest) value of the expression is \( c \). The last of these is why “find the minimum value” questions never need calculus at P1.

The method, when \( a \ne 1 \): factor \(a\) out of the \( x^{2} \) and \( x \) terms only, complete the square inside the bracket, then multiply back out.

The step people drop is multiplying the \( -9 \) by the 2. If your completed square does not expand back to the original, that is almost always where it went wrong — and expanding back takes ten seconds.

Note the sign convention: \( (x-3)^{2} \) puts the vertex at \( x = +3 \). Whatever makes the bracket zero is where the vertex sits.

The discriminant

For \( ax^{2} + bx + c = 0 \), the discriminant is \( b^{2} - 4ac \), and it tells you about the roots without finding them:

Translate the geometry into the algebra and back, because questions are phrased either way:

• “cuts the \(x\)-axis twice”, “meets the curve at two points” → \( > 0 \)
• “touches”, “is a tangent to”, “has equal roots” → \( = 0 \)
• “does not meet”, “is always positive” (with \(a > 0\)) → \( < 0 \)

Questions asking for the values of a constant \(k\) that produce a given number of roots are discriminant questions, and they almost always end in a quadratic inequality in \(k\).

Quadratic inequalities

To solve \( ax^{2} + bx + c > 0 \): find the critical values by solving the corresponding equation, sketch the parabola, and read off the region. Do not divide an inequality by an expression whose sign you do not know, and do not simply reverse the sign of the equation’s answer.

The shape decides the form of the answer, and this is where marks are lost:

• Positive \(a\), and you want \( > 0 \) — the curve is above the axis outside the roots, so the answer is two separate regions: \( x < \alpha \) or \( x > \beta \).
• Positive \(a\), and you want \( < 0 \) — the curve is below the axis between the roots, so the answer is a single interval: \( \alpha < x < \beta \).

Disguised quadratics

Many equations become quadratic after a substitution. Spot the pattern: the power of one term is exactly twice the power of another.

Solve for \(u\), then undo the substitution — the answer to the question is in \(x\), not \(u\). Two checks matter at that stage: \( u = x^{2} \) gives two values of \(x\) for each positive \(u\) and none for a negative one; \( u = \sqrt{x} \) requires \( u \ge 0 \), so a negative root of the quadratic must be discarded.

Simultaneous equations: one linear, one quadratic

Make one variable the subject of the linear equation and substitute into the quadratic. That direction round is always easier, and substituting the quadratic into the linear one produces a mess.

The resulting quadratic tells you how the line and the curve are related, which is the discriminant again: two solutions means the line is a chord cutting the curve twice, one repeated solution means it is a tangent, and no real solutions means they never meet.