Shared by a Veda teacher · Generated with Veda AI · Oct 2026
Veda Bites are swipeable micro-lessons — each one teaches exactly one idea. Here's a taste from this kit; the app has the full deck.
Three consecutive squares, one rule, and the answer is always 2.
$1 + 9 - 2(4) = 2$. And $25 + 49 - 2(36) = 2$. And so on, for ever.
↳ Replacing "it keeps happening" with "it must happen" is what this chapter does.
What separates an identity from an equation?
$x + 3 = 7$ is an equation: true for one value of $x$ and false for the rest.
↳ An equation has something to solve; an identity is a tool to use.
The first identity, and the picture behind it.
$$(a + b)^2 = a^2 + 2ab + b^2$$
↳ With $a = 3$ and $b = 2$: $9 + 6 + 6 + 4 = 25 = 5^2$.
Can you lay out $x^2 + 5x + 6$ as a rectangle with no gaps?
Take one $x$ by $x$ square, five $x$ by 1 strips and six unit squares, and fit them into a single rectangle.
↳ Algebra tiles turn factorising from guesswork into building a shape.
Why do the numbers 2 and 3 appear in the answer?
They multiply to 6, because they form the block of six unit squares.
↳ That double role is the whole method, tiles or no tiles.
The identity the picture is showing.
$$(x + a)(x + b) = x^2 + (a + b)x + ab$$
↳ The coefficient of $x$ is the sum, and the constant is the product.
An identity is an equation true for every value of its letters, and the square identities can all be read off a picture: cut a square of side a + b and the four pieces give a squared plus two ab rectangles plus b squared…
~40 min · full explanation, examples & memory tricks in the app
Lay the tiles of a quadratic out as a rectangle and its two side lengths are its factors. Without tiles the same thing is done by finding two numbers that multiply to the constant and add to the middle coefficient, and w…
~45 min · full explanation, examples & memory tricks in the app
Multiplying a square by one more copy of its bracket produces the cube identities, and two of them factorise: unlike a sum of two squares, a sum of two cubes breaks into brackets. Those factorisations, together with ever…
~45 min · full explanation, examples & memory tricks in the app
174 questions laddered from warm-up to topper-level, each with an explanation. A taste:
Expand $(x + 6)^2$.
Beginner$x^2 + 12x + 36$
Using $(a+b)^2 = a^2 + 2ab + b^2$ with $b = 6$ gives a middle term of $2 \times x \times 6 = 12x$.
Expand $(x - 9)^2$.
Beginner$x^2 - 18x + 81$
The middle term is $-2 \times x \times 9$, and the last is $(-9)^2 = 81$, which is positive.
Expand $(2x + 5)^2$.
Intermediate$4x^2 + 20x + 25$
Here $a = 2x$, so $a^2 = 4x^2$ and $2ab = 2(2x)(5) = 20x$.
Expand $(3x - 1)^2$.
Intermediate$9x^2 - 6x + 1$
$a = 3x$ gives $9x^2$, the middle is $-2(3x)(1) = -6x$, and the last is $(-1)^2 = 1$.
60 flashcards in this kit — the app reviews them with spaced repetition so the right card returns on the right day.
Every game is built from this kit's own content — scores feed your mastery, so playing counts as studying.
The full Veda Bites deck, complete notes, spaced-repetition flashcards, leveled MCQs, tests and games for this kit — plus Daily Facts and the Arena, every day.
