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1.1 Quadratics

Syllabus
9709–2028–2029
Topic
1.1
Level
AS

Learning objectives

1.1.1Completing the square• carry out the process of completing the square for a quadratic polynomial ax2 + bx + c and use a completed square form e.g. to locate the vertex of the graph of y = ax2 + bx + c or to sketch the graph.1.1.2Quadratic discriminant• find the discriminant of a quadratic polynomial ax2 + bx + c and use the discriminant e.g. to determine the number of real roots of the equation ax2 + bx + c = 0. Knowledge of the term 'repeated root' is included.1.1.3Quadratic equations• solve quadratic equations, and quadratic inequalities, in one unknown By factorising, completing the square and using the quadratic formula.1.1.4Simultaneous equations• solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic e.g. x + y + 1 = 0 and x2 + y2 = 25, 2x + 3y = 7 and 3x2 = 4 + 4xy.1.1.5Quadratic-form equations• recognise and solve equations in x which are quadratic in some function of x. e.g. x4 - 5x2 + 4 = 0, xx61 0-+ =, tan2 x = 1 + tan x.

Completing the square reveals a quadratic’s turning point and range

Rewrite ax²+bx+c as a(x−h)²+k by factoring a and adding and subtracting the required square. The vertex is (h,k), and the sign of a tells whether it is a minimum or maximum.

Keep the factor outside the square: ax²+bx+c=a[x²+(b/a)x]+c. Completing the square is also the quickest route to a quadratic equation’s exact roots and range.

x²−6x+5=(x−3)²−4, so the graph has minimum −4 at x=3 and range y≥−4.

The constant changes when the square is completed; do not write (x−3)²+5, which expands to a different quadratic.

The discriminant classifies real roots before you solve the quadratic

For ax²+bx+c=0, the discriminant Δ=b²−4ac. Δ>0 gives two distinct real roots, Δ=0 one repeated real root, and Δ<0 no real roots.

Use the discriminant when the question asks how many roots, a tangent condition or a parameter range. It is often cleaner than applying the quadratic formula twice.

For x²−4x+k=0, Δ=16−4k. Two real roots require k<4; tangency occurs at k=4.

Δ>0 does not mean both roots are positive; root signs require additional information such as sum and product.

Choose a quadratic method, then turn inequality roots into sign intervals

For ax2+bx+c=0ax^2+bx+c=0, factorise when simple factors are visible, complete the square when vertex/range structure matters, or use the quadratic formula for a general or parameterised quadratic.

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Example: 2x2+3x−2=02x^2+3x-2=0 gives (2x−1)(x+2)=0(2x-1)(x+2)=0, so x=12x=\tfrac12 or x=−2x=-2. Keep the ±\pm branch when square roots arise and give exact roots unless approximation is requested.

For a quadratic inequality, first find the boundary roots, place them on a number line, then determine the sign in each interval from the parabola or a test value. Include a root only for ≤\leq or ≥\geq.

$2x^2+3x-2<0$ has roots $-2$ and $\tfrac12$. Since the leading coefficient is positive, the quadratic is negative between them:-2<x<\tfrac12.

Roots solve the boundary equation; they are not automatically the answer to an inequality. Report intervals with the correct strict or inclusive endpoints.

Substitute the linear equation, solve every branch, then rebuild coordinate pairs

When one simultaneous equation is linear and the other quadratic, rearrange the linear equation for one variable and substitute it into the quadratic. Solve the resulting quadratic, back-substitute every valid root and check each ordered pair in both originals.

2x+y+4=0,\qquad 2xy+5y^2=24From the line, $x=-(y+4)/2$.

2\left(-\frac{y+4}{2}\right)y+5y^2=244y^2-4y-24=0;\Rightarrow;(y-3)(y+2)=0

Thus y=3y=3 gives x=−72x=-\tfrac72, and y=−2y=-2 gives x=−1x=-1. The intersection points are (−72,3)(-\tfrac72,3) and (−1,−2)(-1,-2); substitution in both originals verifies them.

Algebraic solution pairs are exactly the graph intersections. Zero, one or two real pairs correspond to no intersection, tangency or two intersections in the usual line-quadratic case.

A root for yy is not a complete solution. Pair it with its own back-substituted xx value; do not mix coordinates from different branches.

Treat the repeated expression as one variable, then filter before reversing

An equation is quadratic in a repeated expression g(x)g(x) when it can be written A[g(x)]2+B[g(x)]+C=0A[g(x)]^2+B[g(x)]+C=0. Set u=g(x)u=g(x), solve the quadratic in uu, then solve g(x)=ug(x)=u for every admissible value.

Use four checks: identify the same repeated expression, substitute without changing coefficients, apply the range/domain of gg, then reverse the substitution completely and verify in the original equation.

x^4-5x^2+4=0Set $u=x^2$: $(u-1)(u-4)=0$, so $u=1$ or $4$. Both satisfy $u\geq0$, hencex=\pm1,;\pm2.

The same structure may use u=x3u=x^3, u=1/xu=1/x, u=tan⁡xu=\tan x or a stated expression such as u=2x−3u=2x-3. Its restrictions differ: x2≥0x^2\geq0, 1/x≠01/x\ne0, and trigonometric reversal must respect the required interval.

A valid quadratic root for uu can still be impossible for g(x)g(x). Reject it only from the actual range/domain, and do not forget multiple xx values when reversing squares or trigonometric functions.