Question: Show that $x^2 - 4x + 7 > 0$ for all real values of $x$.

Solution:

Complete the square:

$$\begin{aligned} x^2 - 4x + 7 &= (x - 2)^2 - 4 + 7 \\ &= (x - 2)^2 + 3 \end{aligned}$$

Since $(x - 2)^2 \geq 0$ for all real $x$ (a square is never negative),

$$(x - 2)^2 + 3 \geq 0 + 3 = 3 > 0$$

Therefore $x^2 - 4x + 7 > 0$ for all real $x$. (shown)

Geometric reading: the parabola has minimum value $3$ at $x = 2$, so the whole curve lies above the $x$-axis.

Question: Show that $4x^2 - 6x + 14 \geq \tfrac{47}{4}$ for all real $x$.

Solution:

$$\begin{aligned} 4x^2 - 6x + 14 &= 4\left(x^2 - \tfrac{3}{2}x\right) + 14 \\ &= 4\left[\left(x - \tfrac{3}{4}\right)^2 - \tfrac{9}{16}\right] + 14 \\ &= 4\left(x - \tfrac{3}{4}\right)^2 - \tfrac{9}{4} + 14 \\ &= 4\left(x - \tfrac{3}{4}\right)^2 + \tfrac{47}{4} \end{aligned}$$

Since $4\left(x - \tfrac{3}{4}\right)^2 \geq 0$,

$$4x^2 - 6x + 14 \geq \tfrac{47}{4}$$

Minimum value is $\tfrac{47}{4}$, achieved at $x = \tfrac{3}{4}$. (shown)

Question: The roots of $3x^2 + 4x - 2 = 0$ are $\alpha$ and $\beta$. Find $\alpha + \beta$ and $\alpha\beta$ without solving the equation.

Key formulas:

$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

Solution:

Read coefficients: $a = 3$, $b = 4$, $c = -2$.

$$\alpha + \beta = -\frac{4}{3}, \qquad \alpha\beta = \frac{-2}{3} = -\frac{2}{3}$$

⚠️ Sign trap: the minus is only on the sum formula, never on the product. And $b$, $c$ keep their original signs — for $6x^2 - x - 3 = 0$ you'd get $\alpha + \beta = -\tfrac{-1}{6} = \tfrac{1}{6}$ (double-negative becomes positive).

Question: If $\alpha, \beta$ are the roots of $4x^2 + 7x + 1 = 0$, find $\alpha^2 + \beta^2$.

Key identity:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

Solution:

From the equation: $\alpha + \beta = -\tfrac{7}{4}$, $\alpha\beta = \tfrac{1}{4}$.

$$\begin{aligned} \alpha^2 + \beta^2 &= \left(-\tfrac{7}{4}\right)^2 - 2\left(\tfrac{1}{4}\right) \\ &= \tfrac{49}{16} - \tfrac{1}{2} \\ &= \tfrac{49}{16} - \tfrac{8}{16} \\ &= \boxed{\tfrac{41}{16}} \end{aligned}$$

Why this works: expanding $(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ and rearranging gives the identity. So you never need to solve the equation.

Question: If $\alpha, \beta$ are the roots of $x^2 + 4x + 5 = 0$, find $\alpha^3 + \beta^3$.

Key identity:

$$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$

Solution:

From the equation: $\alpha + \beta = -4$, $\alpha\beta = 5$.

$$\alpha^3 + \beta^3 = (-4)^3 - 3(5)(-4)$$ $$= -64 + 60 = \boxed{-4}$$

For higher powers like $\alpha^4 + \beta^4$, use $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$ — same chain of substitutions.

Question: The roots of $3x^2 - 2x + 5 = 0$ are $\alpha, \beta$. Find the quadratic equation whose roots are $\tfrac{1}{\alpha}$ and $\tfrac{1}{\beta}$.

Key formula: A quadratic with roots $r_1, r_2$ is

$$x^2 - (\text{SOR})\,x + (\text{POR}) = 0$$

Solution:

From original: $\alpha + \beta = \tfrac{2}{3}$, $\alpha\beta = \tfrac{5}{3}$.

New sum of roots:

$$\tfrac{1}{\alpha} + \tfrac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{2/3}{5/3} = \tfrac{2}{5}$$

New product of roots:

$$\tfrac{1}{\alpha} \cdot \tfrac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{5/3} = \tfrac{3}{5}$$

New equation:

$$x^2 - \tfrac{2}{5}x + \tfrac{3}{5} = 0 \quad\text{or}\quad \boxed{5x^2 - 2x + 3 = 0}$$

Question: If $\alpha, \beta$ are roots of $2x^2 - 7x + 8 = 0$, form the equation whose roots are $\alpha^2 - 1$ and $\beta^2 - 1$.

Solution:

From original: $\alpha + \beta = \tfrac{7}{2}$, $\alpha\beta = 4$.

New sum of roots:

$$\begin{aligned} (\alpha^2 - 1) + (\beta^2 - 1) &= (\alpha^2 + \beta^2) - 2 \\ &= (\alpha+\beta)^2 - 2\alpha\beta - 2 \\ &= \left(\tfrac{7}{2}\right)^2 - 2(4) - 2 \\ &= \tfrac{49}{4} - 10 = \tfrac{9}{4} \end{aligned}$$

New product of roots:

$$\begin{aligned} (\alpha^2 - 1)(\beta^2 - 1) &= \alpha^2\beta^2 - (\alpha^2 + \beta^2) + 1 \\ &= (\alpha\beta)^2 - \left[(\alpha+\beta)^2 - 2\alpha\beta\right] + 1 \\ &= 16 - \left[\tfrac{49}{4} - 8\right] + 1 \\ &= 16 - \tfrac{17}{4} + 1 = \tfrac{51}{4} \end{aligned}$$

New equation:

$$x^2 - \tfrac{9}{4}x + \tfrac{51}{4} = 0 \implies \boxed{4x^2 - 9x + 51 = 0}$$

Question: The roots of $x^2 - 3x + k = 0$ differ by $3$. Find $k$.

Strategy: Parameterise the roots using the relationship, then equate to the sum/product formulas.

Solution:

Let the roots be $\alpha$ and $\alpha - 3$.

Sum of roots:

$$\alpha + (\alpha - 3) = -\tfrac{-3}{1} = 3$$ $$2\alpha - 3 = 3 \implies \alpha = 3$$

Product of roots:

$$\alpha(\alpha - 3) = \tfrac{k}{1} = k$$ $$3 \cdot 0 = k \implies \boxed{k = 0}$$

Question: Find the range of values of $k$ for which $y = x^2 + kx + 9$ has no real roots.

Solution:

"No real roots" means the discriminant is negative:

$$b^2 - 4ac < 0$$

Here $a = 1$, $b = k$, $c = 9$:

$$k^2 - 4(1)(9) < 0$$ $$k^2 - 36 < 0$$ $$(k - 6)(k + 6) < 0$$

Solving the quadratic inequality:

$$\boxed{-6 < k < 6}$$

Question: Find the value(s) of $k$ for which the line $y = 2x + k$ is a tangent to the curve $y = x^2 - 4x + 5$.

Solution:

Step 1. Substitute the line into the curve to get a single quadratic in $x$:

$$2x + k = x^2 - 4x + 5$$ $$x^2 - 6x + (5 - k) = 0$$

Step 2. "Tangent" means exactly one point of intersection → discriminant $= 0$:

$$b^2 - 4ac = 0$$ $$(-6)^2 - 4(1)(5 - k) = 0$$ $$36 - 20 + 4k = 0$$ $$4k = -16 \implies \boxed{k = -4}$$

Question: Find the points of intersection between the line $y = 5 - 3x$ and the curve $y = -x^2 + 4x - 7$.

Solution:

Step 1. Equate the two $y$ expressions:

$$5 - 3x = -x^2 + 4x - 7$$ $$x^2 - 7x + 12 = 0$$

Step 2. Factorise:

$$(x - 3)(x - 4) = 0 \implies x = 3 \text{ or } x = 4$$

Step 3. Get $y$ from the easier equation $y = 5 - 3x$:

At $x = 3$: $y = 5 - 9 = -4$ → point $A(3, -4)$.

At $x = 4$: $y = 5 - 12 = -7$ → point $B(4, -7)$.

Intersection points: $\boxed{(3, -4) \text{ and } (4, -7)}$.

Question: Find the values of $k$ for which the line $2y = 6x + k$ intersects the curve $y = x^2 - 5x + 5$ at two distinct points.

Solution:

Rewrite the line: $y = 3x + \tfrac{k}{2}$. Substitute into the curve:

$$3x + \tfrac{k}{2} = x^2 - 5x + 5$$ $$x^2 - 8x + \left(5 - \tfrac{k}{2}\right) = 0$$

Two distinct points → discriminant $> 0$:

$$(-8)^2 - 4(1)\left(5 - \tfrac{k}{2}\right) > 0$$ $$64 - 20 + 2k > 0$$ $$2k > -44 \implies \boxed{k > -22}$$

Question: A stone is thrown vertically upwards. Its height $h$ metres at time $t$ seconds is $h = 16t - 4t^2 + 1$.

(a) State the initial height. (b) Find the maximum height.

Solution:

(a) Initial height = height at $t = 0$:

$$h = 16(0) - 4(0)^2 + 1 = 1 \text{ m}$$

(b) Complete the square to find the maximum:

$$\begin{aligned} h &= -4t^2 + 16t + 1 \\ &= -4(t^2 - 4t) + 1 \\ &= -4\left[(t - 2)^2 - 4\right] + 1 \\ &= -4(t - 2)^2 + 17 \end{aligned}$$

Since coefficient is $-4 < 0$, max occurs at $t = 2$ s. Maximum height = 17 m.

Question: An arched tunnel is modelled by $y = -1.76x^2 + 4.4x$, with $y$ as height and $x$ as horizontal distance (m) from one wall.

(a) Find the highest point of the arch. (b) Can a lorry $1.8$ m wide and $2$ m tall pass through (driving along the centre)?

Solution:

(a) Complete the square:

$$\begin{aligned} y &= -1.76(x^2 - 2.5x) \\ &= -1.76\left[(x - 1.25)^2 - 1.5625\right] \\ &= -1.76(x - 1.25)^2 + 2.75 \end{aligned}$$

Highest point: $(1.25, 2.75)$, i.e. $1.25$ m from one wall, height $2.75$ m.

(b) Lorry $1.8$ m wide, centred on $x = 1.25$, so its sides are at $x = 1.25 \pm 0.9$.

Check height at $x = 0.35$:

$$\begin{aligned} y &= -1.76(0.35)^2 + 4.4(0.35) \\ &= -0.2156 + 1.54 \\ &= 1.3244 \text{ m} \end{aligned}$$

Tunnel is only $1.32$ m tall there, but the lorry is $2$ m tall. The lorry cannot pass.

Question: The curve $y = x^2 - 4x - 5$ is reflected in the $x$-axis. Write the equation of the reflected curve in completed square form, and state its turning point.

Solution:

Reflection in $x$-axis: replace $y$ with $-y$ (or equivalently, multiply the right-hand side by $-1$):

$$\begin{aligned} -y &= x^2 - 4x - 5 \\ y &= -(x^2 - 4x - 5) \\ &= -x^2 + 4x + 5 \end{aligned}$$

Complete the square:

$$\begin{aligned} y &= -(x^2 - 4x) + 5 \\ &= -\left[(x - 2)^2 - 4\right] + 5 \\ &= -(x - 2)^2 + 9 \end{aligned}$$

Since coefficient is $-1 < 0$, the reflected curve has a maximum point at $(2, 9)$.

Sanity check: the original $y = x^2 - 4x - 5 = (x-2)^2 - 9$ has minimum at $(2, -9)$. Reflecting in the $x$-axis flips the $y$-coordinate's sign: $(2, -9) \to (2, 9)$. ✓

Question: A quadratic curve has $x$-intercepts at $x = -2$ and $x = p$. The axis of symmetry is $x = 3$. Find $p$, and write the equation of the curve in factored form (with leading coefficient $1$).

Solution:

The axis of symmetry of a quadratic passes through the midpoint of its $x$-intercepts:

$$\frac{-2 + p}{2} = 3$$ $$-2 + p = 6 \implies \boxed{p = 8}$$

With roots $-2$ and $8$ and leading coefficient $1$:

$$y = (x + 2)(x - 8)$$

Useful fact: the axis of symmetry of $y = ax^2 + bx + c$ is at $x = -\tfrac{b}{2a}$, which is also the average of the two roots.

Question:

(a) Express $y_1 = -x^2 + 6x - 14$ and $y_2 = 3x^2 + 12x + 13$ each in completed square form.

(b) Hence explain why $y_1$ and $y_2$ do not intersect.

Solution:

(a)

$$\begin{aligned} y_1 &= -(x^2 - 6x + 14) \\ &= -\left[(x-3)^2 - 9 + 14\right] \\ &= -(x-3)^2 - 5 \end{aligned}$$ $$\begin{aligned} y_2 &= 3(x^2 + 4x) + 13 \\ &= 3\left[(x+2)^2 - 4\right] + 13 \\ &= 3(x+2)^2 + 1 \end{aligned}$$

(b) $y_1$ has maximum value $-5$ (since $a = -1 < 0$), so $y_1 \leq -5$ for all $x$.

$y_2$ has minimum value $1$ (since $a = 3 > 0$), so $y_2 \geq 1$ for all $x$.

Since $y_1 \leq -5 < 1 \leq y_2$ for every $x$, the two curves never share a $y$-value. They do not intersect.

Question: Express $y = x^2 - 4x$ in the form $a(x - h)^2 + k$, and hence state the coordinates of its turning point.

Solution:

Take half the coefficient of $x$ (which is $-4$), square it ($\left(\tfrac{-4}{2}\right)^2 = 4$), then add and subtract:

$$y = x^2 - 4x + 4 - 4 = (x - 2)^2 - 4$$

Comparing with $a(x-h)^2 + k$: $a = 1$, $h = 2$, $k = -4$.

Since $a > 0$, the parabola opens upward → minimum point at $(2, -4)$.

Question: Express $y = 2x^2 - 10x - 18$ in completed square form and find the turning point.

Solution:

Step 1. Factor out $2$ from the $x^2$ and $x$ terms only:

$$y = 2(x^2 - 5x) - 18$$

Step 2. Inside the bracket, half of $-5$ is $-\tfrac{5}{2}$, square it: $\tfrac{25}{4}$.

$$\begin{aligned} y &= 2\left[x^2 - 5x + \tfrac{25}{4} - \tfrac{25}{4}\right] - 18 \\ &= 2\left[\left(x - \tfrac{5}{2}\right)^2 - \tfrac{25}{4}\right] - 18 \end{aligned}$$

Step 3. Distribute the $2$:

$$\begin{aligned} y &= 2\left(x - \tfrac{5}{2}\right)^2 - \tfrac{25}{2} - 18 \\ &= 2\left(x - \tfrac{5}{2}\right)^2 - \tfrac{61}{2} \end{aligned}$$

Since $a = 2 > 0$ → minimum point at $\left(\tfrac{5}{2}, -\tfrac{61}{2}\right)$.

Question: Express $y = -4x^2 - 24x - 35$ in completed square form. Hence sketch the curve, labelling the turning point and intercepts.

Solution:

$$\begin{aligned} y &= -4\left(x^2 + 6x + \tfrac{35}{4}\right) \\ &= -4\left[(x+3)^2 - 9 + \tfrac{35}{4}\right] \\ &= -4(x + 3)^2 + 1 \end{aligned}$$

Turning point: $(-3, 1)$, maximum since $a = -4 < 0$.

$x$-intercepts: Set $y = 0$:

$$\begin{aligned} -4(x+3)^2 + 1 &= 0 \\ (x+3)^2 &= \tfrac{1}{4} \\ x &= -\tfrac{7}{2} \text{ or } -\tfrac{5}{2} \end{aligned}$$

$y$-intercept: $y = -4(0+3)^2 + 1 = -35$, point $(0, -35)$.

Sketch: downward parabola, vertex at $(-3, 1)$, $x$-intercepts $-\tfrac{7}{2}$ and $-\tfrac{5}{2}$, $y$-intercept $-35$.

Question: The curve $y = (x-2)^2 - 4$ has a minimum point. State the minimum value of $y$.

Solution:

Turning point is $(2, -4)$. The minimum value of $y$ is the $y$-coordinate of the turning point:

$$\boxed{y_{\min} = -4}$$

⚠️ Common mistake: Don't write "minimum value = 2" — that's the $x$-coordinate. The min/max value always refers to $y$, not $x$. Visual check: the lowest point of the parabola is $-4$ on the $y$-axis, not $2$.

Question: $\alpha$ and $\beta$ are the roots of $x^2 - 5x + 6 = 0$. Find a quadratic equation whose roots are $\alpha + \beta$ and $\alpha\beta$.

Solution:

From the original: $\alpha + \beta = 5$ and $\alpha\beta = 6$.

So the new roots are simply $5$ and $6$.

New sum of roots: $5 + 6 = 11$.

New product of roots: $5 \times 6 = 30$.

Construct using $x^2 - (\text{SOR})\,x + (\text{POR}) = 0$:

$$\boxed{x^2 - 11x + 30 = 0}$$

General pattern: any symmetric expression in $\alpha, \beta$ — e.g. $\alpha^2 + \beta^2$, $\tfrac{1}{\alpha} + \tfrac{1}{\beta}$, $\alpha\beta + \alpha + \beta$ — can be evaluated using just $\alpha + \beta$ and $\alpha\beta$.