x + y = 10 and xy = 21. What is x² + y²?

Core

x + y = 10, x y = 21 x^{2} + y^{2} = ?

Transcript

Two numbers add to $10$ and multiply to $21$ . What is $x^{2} + y^{2}$ ? Most people say $100$ .

$x + y = 10 x y = 21$

$x^{2} + y^{2} = ?$

$x^{2} + y^{2} = 100$

Don't hunt for the numbers. Square the sum: $(x + y)^{2} = 1 0^{2} = 100$ .

$(x + y)^{2} = 1 0^{2} = 100$

A square with side $x + y$ . Cut each side where $x$ ends and $y$ begins.

[A square whose side is x + y, cut into four cells: an x-by-x square, a y-by-y square, and two identical x-by-y strips between them.]

[x + y]

[Two cuts, one down and one across, made where x ends and y begins.]

Four pieces: $x^{2}$ , $y^{2}$ — and two strips, each $x \cdot y$ .

[x²]

[y²]

[xy]

Each strip is $21$ . Two strips: $2 x y = 42$ .

$= 21$

$2 \cdot 21 = 42$

$x^{2} + y^{2} = 100 - 42 = 58$

You never found $x$ or $y$ . A sum and a product build any symmetric answer.

The two strips that squaring forgets · 1:39

The eye wants to square both sides and be done: $x + y = 10$ , so $x^2 + y^2 = 100$ . But squaring a sum is not squaring its pieces. $(x+y)^2$ means $(x+y)(x+y)$ , and that product opens into four terms, not two:

$(x + y)^2 = x^2 + xy + yx + y^2 = x^2 + 2xy + y^2$

Now look at what the problem already told you. It gave you the left side — $(x+y)^2 = 10^2 = 100$ — and it gave you the middle, $xy = 21$ , so the two extra terms are worth $2 \times 21 = 42$ . Subtract exactly what the square added:

$x^2 + y^2 = (x + y)^2 - 2xy = 100 - 42 = 58$

Notice what never happened: you never found $x$ or $y$ . Here you could have — “two numbers with sum $10$ and product $21$ ” is M2’s hunt, and it turns up $3$ and $7$ , giving $9 + 49 = 58$ . Same answer, longer road. Change the product to $23$ and that road closes: no nice pair exists, $x$ and $y$ are $5 \pm \sqrt{2}$ , and the one-line route still says $100 - 46 = 54$ without blinking.

What this hides

The trap isn’t carelessness — it’s a real rule misfiring. Multiplication does distribute over addition, so “square each piece” feels like the same move. It isn’t: squaring is a product of two parentheses, and every term of the first meets every term of the second. The two middle cells are the entire difference between $100$ and $58$ , and they’re invisible only because the shortcut never draws them.

The transferable idea is bigger than the identity. $x^2 + y^2$ is symmetric — swap $x$ and $y$ and nothing changes — and every symmetric combination of two unknowns can be built out of just their sum and their product. So when a question hands you $x + y$ and $xy$ and asks for something symmetric, that’s not a missing piece of information: it’s the whole toolkit. Don’t solve for $x$ and $y$ ; assemble the answer from the two numbers you were given.

Watch the missing rectangles exist. Load the $(x+4)(x+4)$ preset — a square, same shape as $(x+y)^2$ — and count the cells: one $x^2$ , one constant, and two middle strips that the “square each piece” shortcut throws away.

The square that forgets two cells