Write 9x^2 - 12xy + 4y^2 as a perfect square.

Step-by-step solution

1. 1

   Test whether the two end terms are perfect squares. A trinomial can only be a squared binomial if its first and last terms are squares:

   $9x^2=(3x)^2,\qquad 4y^2=(2y)^2$

   So the candidate is $(3x\pm 2y)^2$, and only the middle term can decide the sign.

2. 2

   Check the middle term against $2ab$. For $(a-b)^2=a^2-2ab+b^2$ with $a=3x$ and $b=2y$:

   $2ab=2(3x)(2y)=12xy$

   The given middle term is $-12xy$, which matches $-2ab$ exactly. This test is the whole method — if the middle term had been anything other than $\pm 12xy$, the expression would not be a perfect square.

3. 3

   Choose the sign from the middle term. Since the middle term is negative, the binomial carries a minus:

   $9x^2-12xy+4y^2=(3x-2y)^2$

4. 4

   Expand to confirm.

   $(3x-2y)^2=(3x)^2-2(3x)(2y)+(2y)^2=9x^2-12xy+4y^2\;\checkmark$

5. 5

   Note what the factored form tells you. Because the expression is a square, it is never negative for real $x,y$, and it equals zero exactly when $3x=2y$. Substituting $x=2$, $y=3$ gives $9(4)-12(6)+4(9)=36-72+36=0$, and indeed $3(2)=2(3)\;\checkmark$ — a numeric confirmation of the zero set.