Show that $x^2 - 10x + 21$ can never be smaller than $-4$. Given that $y = x^2 - 10x + 21$, write down the coordinates of the turning point and state whether the point is a maximum or minimum. Solve the equation $x^2 - 10x + 21 = 0$. Hence, sketch the graph of $y = x^2 - 10x + 21$. Given that $6x^2 + \frac{x}{a} - \frac{1}{b^2} = 0$, where $a, b \ne 0$, show that there are no values of $a$ and $b$ for which the equation has equal roots.

1. Complete the square for the quadratic expression

$x^2 - 10x + 21 = (x - 5)^2 - (-5)^2 + 21$
$= (x - 5)^2 - 25 + 21$
$= (x - 5)^2 - 4$

2. Use the property of squares to show the minimum value

For any real number $x$, $(x - 5)^2 \geqslant 0$.
Therefore, $(x - 5)^2 - 4 \geqslant -4$.
Hence, $x^2 - 10x + 21$ can never be smaller than $-4$.

1. Deduce the coordinates of the turning point using the result from part (a)(i)

From (a)(i), $x^2 - 10x + 21 = (x - 5)^2 - 4$.
The minimum value is $-4$, which occurs when $x - 5 = 0 \implies x = 5$.
Coordinates of the turning point are $(5, -4)$.

2. Determine the nature of the turning point

Since the coefficient of $x^2$ is $1 > 0$, the curve opens upwards.
Therefore, the turning point is a minimum.

1. Factorise the quadratic expression

$x^2 - 10x + 21 = 0$
$(x - 3)(x - 7) = 0$

2. Solve for x

$x - 3 = 0 \implies x = 3$
$x - 7 = 0 \implies x = 7$

1. Identify the key features of the graph from earlier parts

From (ii), the minimum turning point is $(5, -4)$.
From (iii), the $x$-intercepts are $x = 3$ and $x = 7$.
The $y$-intercept is found by setting $x = 0$: $y = 21$.

2. Sketch the graph using the identified features

Draw a smooth U-shaped curve (since the coefficient of $x^2$ is positive) passing through $(0, 21)$, $(3, 0)$, and $(7, 0)$, with its lowest point at $(5, -4)$.

1. Formulate the discriminant of the given quadratic equation.

For the equation $6x^2 + \frac{x}{a} - \frac{1}{b^2} = 0$, the discriminant is:
$\Delta = \left(\frac{1}{a}\right)^2 - 4(6)\left(-\frac{1}{b^2}\right)$
$\Delta = \frac{1}{a^2} + \frac{24}{b^2}$

2. Analyse the sign of the discriminant to show it cannot be zero.

Since $a \ne 0$ and $b \ne 0$, their squares are strictly positive: $a^2 > 0$ and $b^2 > 0$.
Therefore, $\frac{1}{a^2} > 0$ and $\frac{24}{b^2} > 0$.
This means $\Delta = \frac{1}{a^2} + \frac{24}{b^2} > 0$.
For the equation to have equal roots, $\Delta$ must be $0$. Since $\Delta > 0$ for all non-zero $a$ and $b$, $\Delta$ can never be zero. Hence, there are no values of $a$ and $b$ for which the equation has equal roots.