Test Overview and Completing the Square

This test focuses on seven key quadratic skills that form the backbone of A-level algebra. Completing the square transforms expressions like $5x^2 + 10x + 2$into the more useful form$5$x + 1$^2 - 3$, making it easier to find turning points and solve equations.

The discriminant $b^2 - 4ac$ tells you about the nature of roots. When it equals zero, you get equal roots - this is crucial for finding when equations have exactly one solution. For example, $kx^2 - 30x + 25k = 0$ has equal roots when $k = ±3$.

Real-world modelling appears through projectile motion problems. The arrow's trajectory $H = 1.8 + 0.4d - 0.002d^2$ shows how quadratics describe parabolic paths, where the constant 1.8 represents the initial height above ground.

Key Tip: Always check your completed square form by expanding it back out - this catches sign errors that can cost you marks.

Advanced Modelling and Simultaneous Equations

Simultaneous equations involving quadratics create intersection problems between curves and lines. When solving $y = x^2 - 5x + 4$ and $y = x - 1$, you're finding where a parabola meets a straight line. The number of solutions tells you how many intersection points exist.

Tangent conditions occur when the discriminant equals zero. For a line $y = x + c$ to touch (be tangent to) a curve, there must be exactly one intersection point. This gives you the condition to find the value of $c$.

Projectile modelling extends to golf balls and other real situations. The trajectory passing through $(0,0)$, $(50,20)$, and $(100,0)$ creates the equation $H = -\frac{1}{125}(x - 50)^2 + 20$, showing the classic parabolic flight path.

Model limitations are equally important - assuming flat ground, ignoring air resistance, or neglecting spin effects are typical restrictions you should always consider.

Key Tip: When finding tangent conditions, remember that tangency means the discriminant equals zero - one solution only.

Complex Simultaneous Systems and Function Analysis

Non-linear simultaneous equations like $y + 4x + 1 = 0$ and $y^2 + 5x^2 + 2x = 0$ require substitution techniques. Replace the linear equation into the quadratic to create a solvable form, then work backwards to find both variables.

Disguised quadratics appear in equations like $2x^4 - 15x^2 - 50 = 0$. By substituting$a = x^2$, this becomes$2a^2 - 15a - 50 = 0$, which factors as$$2a + 5$$a - 10$ = 0$. Remember to convert back to find the original variable.

Sketching quadratic functions requires identifying key features: vertex coordinates, y-intercept, and line of symmetry. For $f(x) = 11 - 4x - 2x^2$, completing the square gives $13 - 2$x + 1$^2$, showing vertex$(-1, 13)$and line of symmetry$x = -1$.

Proving positivity uses completed square form. Since $(x - 3)^2 \geq 0$ for all real $x$, expressions like $x^2 - 6x + 10 = (x - 3)^2 + 1$ are always positive because they equal $(x - 3)^2 + 1 \geq 1 > 0$.

Key Tip: For surd form answers, always rationalise denominators and simplify completely - examiners expect exact forms, not decimals.