Read quadratic roots and signs

Choose a form for the task. Keep exact surds until the final approximation; check solutions in the original equation, particularly after squaring.

Recognise the number systems

Classify a number by whether it can be written as a ratio of integers. A decimal approximation alone does not establish that a number is rational.

The revision starts with the integers $\mathbb Z=\{\ldots,-2,-1,0,1,2,3,4,\ldots\}$ and positive integers $\mathbb Z^+=\{1,2,3,4,\ldots\}$. Rational numbers have the form

$$\mathbb Q=\left\{\frac pq:p,q\in\mathbb Z,\ q\ne0\right\}.$$

The source examples are $-\frac12$, $\frac56$, $\frac83$, $2122$, $12.6$, $-99.4$ and $-0.875$. Integers and finite decimals can also be written as such fractions.

The real numbers $\mathbb R$ are all numbers on the number line; $\mathbb R^+=\{x\in\mathbb R:x>0\}$. An irrational number is a real number that is not rational. The printed examples are $\sqrt3\approx1.73205\ldots$, $\pi\approx3.14159\ldots$, $e\approx2.71828\ldots$ and $\ln2\approx0.693147\ldots$.

Complete a square without changing the polynomial

Halve the coefficient of x, square it, then add and subtract that same amount. Expand your result to check that you have preserved the polynomial.

A quadratic polynomial is $ax^2+bx+c$ with constant coefficients and $a\ne0$. Completing the square writes it as $a(x+p)^2+q$. The source identity is

$$x^2+nx=\left(x+\frac n2\right)^2-\left(\frac n2\right)^2.$$

For example, $x^2+4x=(x+\frac42)^2-(\frac42)^2=(x+2)^2-4$. When the leading coefficient is not 1, first factor it from the variable terms, then complete the square inside the bracket; every correction is still multiplied by that coefficient. Example 1 retains both printed calculations. Continue with Tutorial Question 2.

Derive the quadratic formula

The formula comes from completing the square. Follow the rearrangement in two stages: isolate a square, then take both square-root signs.

Starting from $ax^2+bx+c=0$, with $a\ne0$, the printed derivation successively gives

$$\begin{aligned} a\left(x^2+\frac ba x\right)+c&=0,\\ a\left(x^2+\frac ba x\right)&=-c,\\ x^2+\frac ba x&=-\frac ca,\\ \left(x+\frac b{2a}\right)^2-\left(\frac b{2a}\right)^2&=-\frac ca,\\ \left(x+\frac b{2a}\right)^2&=-\frac ca+\frac{b^2}{4a^2}. \end{aligned}$$

Taking both square-root signs and subtracting $b/(2a)$ gives

$$\begin{aligned} x+\frac b{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}},\\ x&=-\frac b{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}},\\ x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \end{aligned}$$

For real roots the radicand must be nonnegative. The final ± expression represents the same pair for either sign of a; it does not assert $\sqrt{a^2}=a$ for negative a.

The quantity $D=b^2-4ac$ is the discriminant. Example 2 uses this formula with a parameter. Continue with Tutorial Question 3.

Read roots and strict signs from a graph

Ask two questions: which way does the parabola open, and how many times does it meet the x-axis? The leading coefficient answers the first; the discriminant answers the second.

For real coefficients and $a\ne0$, $D>0$ gives two real distinct roots; $D=0$ gives two equal real roots (one distinct root); $D<0$ gives no real roots. The six source sketches combine each case with a>0 or a<0. An upward parabola crosses, touches or stays above the x-axis; a downward parabola crosses, touches or stays below it.

The exploration lets you compare all six cases one at a time. Its labelled equations are illustrative examples, not coordinates prescribed by the source; the original schematic remains linked. Real roots exist exactly when $D\ge0$, equivalently when the graph has a point on the x-axis, including tangency.

A quadratic is strictly positive for every real x exactly when $a>0$ and $D<0$. It is strictly negative for every real x exactly when $a<0$ and $D<0$. Both strict-sign conditions need a negative discriminant: D=0 touches zero and fails strict positivity or negativity. Examples 3–4 retain the source calculations. Continue with Tutorial Question 4 and Tutorial Question 5.