Quadratics

1.1 Quadratics

Completing the square

Completing the square rewrites $ax^2 + bx + c$ as $a(x + p)^2 + q$. This form shows the turning point of the curve straight away.

For $x^2 + bx + c$, halve the coefficient of $x$, then subtract the square of what you halved:

$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$

$x^2 + 6x + 11 = (x + 3)^2 - 9 + 11 = (x + 3)^2 + 2$

When $a \neq 1$, take $a$ out of the first two terms only.

$2x^2 - 8x + 3 = 2(x^2 - 4x) + 3 = 2\left[(x - 2)^2 - 4\right] + 3 = 2(x - 2)^2 - 5$

What the completed square tells you

For $y = a(x + p)^2 + q$, the squared bracket is never negative, so:

So $y = 2(x - 2)^2 - 5$ has a minimum at $(2, -5)$, its line of symmetry is $x = 2$, and its range is $y \geq -5$.

The discriminant

For $ax^2 + bx + c = 0$, the quadratic formula is

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

The part under the root, $b^2 - 4ac$, is the discriminant. It tells you how many real roots there are, and so how the curve meets the $x$-axis.

$k^2 - 4(1)(9) > 0 \quad\Rightarrow\quad k^2 > 36 \quad\Rightarrow\quad k < -6 \text{ or } k > 6$

Quadratic inequalities

Solve the equation first, then use a sketch to pick the correct side.

$(x - 3)(x + 2) > 0$. The U-shaped curve crosses the axis at $x = -2$ and $x = 3$, and is above the axis outside these values:

$x < -2 \quad\text{or}\quad x > 3$

$(2x + 3)(x - 1) \leq 0$. The curve is on or below the axis between the roots:

$-\frac{3}{2} \leq x \leq 1$

Simultaneous equations

When one equation is linear and the other quadratic, make $x$ or $y$ the subject of the linear one and substitute.

Put $y = 5 - x$ into the second equation:

$x^2 + (5 - x)^2 = 13 \quad\Rightarrow\quad 2x^2 - 10x + 12 = 0 \quad\Rightarrow\quad (x - 2)(x - 3) = 0$

So $x = 2, y = 3$ or $x = 3, y = 2$. Always give the pairs together.

Lines and curves

Setting a line equal to a curve gives a quadratic, and its discriminant says how they meet:

$x^2 - 4x + 5 = 2x + k \quad\Rightarrow\quad x^2 - 6x + 5 - k = 0$

For a tangent, $36 - 4(5 - k) = 0$, so $16 + 4k = 0$ and $k = -4$.

Equations that are quadratic in a function of $x$

Some equations are quadratics in disguise. Replace the function with a single letter, solve, then undo the substitution.

Let $u = x^2$: $u^2 - 5u + 4 = 0$, so $u = 1$ or $u = 4$. Then $x^2 = 1$ or $x^2 = 4$, giving $x = \pm 1$ or $x = \pm 2$.

Let $u = \sqrt{x}$: $u^2 - 5u + 6 = 0$, so $u = 2$ or $u = 3$. Then $x = 4$ or $x = 9$.

The same idea handles $2\sin^2 x + \sin x - 1 = 0$ with $u = \sin x$ and $3^{2x} - 4(3^x) + 3 = 0$ with $u = 3^x$.