2.3 Quadratic functions, the discriminant and completing the square — Practise with help

1. $(x + 5)^2 = x^2 + 10x + 25$.
2. $x^2 - 9x + 20 = (x - 4)(x - 5)$.
3. $3^2 - 4 \times 3 + 1 = 9 - 12 + 1 = -2$.

If question 2 was hard, revisit factorising a quadratic with a negative middle term: look for two numbers that multiply to $20$ and add to $-9$.

The coefficient of $x$ is $10$. Half of it is $5$, so the bracket is $(x + 5)^2$.

Expand $(x + 5)^2 = x^2 + 10x + 25$, then subtract the extra $25$ and add the original $3$.

The original has $+3$ where the square has $+25$, so subtract the difference:

Half of $-8$ is $-4$, so the bracket is $(x - 4)^2$. The minimum is the constant left at the end.

Expand $(x - 4)^2 = x^2 - 8x + 16$, then adjust the constant from $16$ down to $1$.

so

A square is never negative, so the minimum value is $-15$, reached when $x = 4$.

The discriminant is $b^2 - 4ac$. Here $a = 1$, $b = 3$ and $c = 7$.

Work out $3^2 - 4 \times 1 \times 7$ and look at its sign.

The discriminant is negative, so the equation has no real roots. The parabola never meets the $x$-axis.

A repeated root means the discriminant is zero. Set $b^2 - 4ac = 0$ with $a = 1$, $b = -12$ and $c = k$.

Solve $144 - 4k = 0$ for $k$.

Set the discriminant to zero:

So

Check: $x^2 - 12x + 36 = (x - 6)^2$, which has the repeated root $x = 6$.

Take the factor $2$ out of the first two terms before completing the square.

Write $2x^2 + 8x + 1 = 2\left(x^2 + 4x\right) + 1$, then complete the square for $x^2 + 4x$.

Take out the factor $2$:

Complete the square inside the bracket:

Substitute back and multiply the $-4$ by $2$:

The vertex is at $(-2, -7)$. Because $a = 2 > 0$, this is a minimum.

"Touches the $x$-axis once" is another way of saying "has a repeated root", so the discriminant is zero.

Solve $(-4)^2 - 4c = 0$ for $c$.

A repeated root needs a zero discriminant:

Then

so the curve touches the $x$-axis at $(2, 0)$.