Sometimes you know the value of x+y or x-y, and of xy, but not the values of x and y themselves. The identities connect these quantities to x2+y2, so you can find it without solving for x and y. **Identity** | **Rearranged** (x+y)2=x2+y2+2xy | x2+y2=(x+y)2-2xy (x-y)2=x2+y2-2xy | x2+y2=(x-y)2+2xy (x+y)2-(x-y)2=4xy | xy= (x+y)2-(x-y)24 (x+y)2+(x-y)2=2(x2+y2) | x2+y2= (x+y)2+(x-y)22 Why: x2+y2, xy and (x+y)2 are linked by one identity, so knowing any two of them fixes the third. The separate values of x and y are never needed. Method - Write down the identity that contains both the given quantities and the one you want. - Substitute the known values. - Solve for the unknown quantity. - If the target has a number in front, such as 5x2+5y2, find x2+y2 first and then multiply. Example Given that x-y=7 and xy=12, find x2+y2 and (x+y)2. x2+y2 = (x-y)2+2xy = 72+2(12) = 73 (x+y)2 = x2+y2+2xy = 73+2(12) = 97 Check: (x+y)2-(x-y)2 should equal 4xy: 97-49=48=4(12). A common error is to square the given value and stop, writing x2+y2=72=49, which forgets the 2xy term.

Finding values with identities

Secondary 2

Sometimes you know the value of $x+y$ or $x-y$, and of $xy$, but not the values of $x$ and $y$ themselves. The identities connect these quantities to $x^2+y^2$, so you can find it without solving for $x$ and $y$.

Identity	Rearranged
$(x+y)^2=x^2+y^2+2xy$	$x^2+y^2=(x+y)^2-2xy$
$(x-y)^2=x^2+y^2-2xy$	$x^2+y^2=(x-y)^2+2xy$
$(x+y)^2-(x-y)^2=4xy$	$xy=\dfrac{(x+y)^2-(x-y)^2}{4}$
$(x+y)^2+(x-y)^2=2(x^2+y^2)$	$x^2+y^2=\dfrac{(x+y)^2+(x-y)^2}{2}$

Why: $x^2+y^2$, $xy$ and $(x+y)^2$ are linked by one identity, so knowing any two of them fixes the third. The separate values of $x$ and $y$ are never needed.

Method

Write down the identity that contains both the given quantities and the one you want.
Substitute the known values.
Solve for the unknown quantity.
If the target has a number in front, such as $5x^2+5y^2$, find $x^2+y^2$ first and then multiply.

Example

Given that $x-y=7$ and $xy=12$, find $x^2+y^2$ and $(x+y)^2$.

$\begin{aligned} x^2+y^2 &= (x-y)^2+2xy \\ &= 7^2+2(12) \\ &= 73 \end{aligned}$

$\begin{aligned} (x+y)^2 &= x^2+y^2+2xy \\ &= 73+2(12) \\ &= 97 \end{aligned}$

Check: $(x+y)^2-(x-y)^2$ should equal $4xy$: $97-49=48=4(12)$. A common error is to square the given value and stop, writing $x^2+y^2=7^2=49$, which forgets the $2xy$ term.

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Identity	Rearranged
\((x+y)^2=x^2+y^2+2xy\)	\(x^2+y^2=(x+y)^2-2xy\)
\((x-y)^2=x^2+y^2-2xy\)	\(x^2+y^2=(x-y)^2+2xy\)
\((x+y)^2-(x-y)^2=4xy\)	\(xy=\dfrac{(x+y)^2-(x-y)^2}{4}\)
\((x+y)^2+(x-y)^2=2(x^2+y^2)\)	\(x^2+y^2=\dfrac{(x+y)^2+(x-y)^2}{2}\)