Squares of Binomials: Additional Practice (Set 2)

The outer roots are $\sqrt{16x^2} = 4x$ and $\sqrt{25y^2} = 5y$. The middle should be $2\cdot 4x\cdot 5y = 40xy$, and here it is $-40xy$, so the sign is negative.

$16x^2 - 40xy + 25y^2 = (4x - 5y)^2$

The negative middle term makes it a difference of the two roots.

For $16x^2 + 20x + 25$ the outer roots are $4x$ and $5$, so a perfect square would need a middle term of $2\cdot 4x\cdot 5 = 40x$, not $20x$.

$2\cdot 4x\cdot 5 = 40x \ne 20x$

The others are $(x + 9)^2$, $(3x - 4)^2$, and $(2x + 1)^2$.

A perfect square needs a middle term equal to $2AB$. For $9x^2 + 15x + 25$ the roots are $3x$ and $5$, so the middle would need to be $2\cdot 3x\cdot 5 = 30x$, not $15x$.

$2\cdot 3x\cdot 5 = 30x \ne 15x$

The others are $(x + 7)^2$, $(2x + 3)^2$, and $(5x - 1)^2$.

Take the first term $5x$ and the last term $3y$.

$(5x + 3y)^2 = (5x)^2 + 2\cdot(5x)\cdot(3y) + (3y)^2 = 25x^2 + 30xy + 9y^2$

The middle term is $30xy$, twice the product $5x\cdot 3y$.

The outer roots are $\sqrt{4x^2} = 2x$ and $\sqrt{9y^2} = 3y$. The middle should be $2\cdot 2x\cdot 3y = 12xy$, and here it is $-12xy$, so the sign is negative.

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

The negative middle term makes it a difference of the roots.

The outer roots are $\sqrt{100x^2} = 10x$ and $\sqrt{1} = 1$. The middle should be $2\cdot 10x\cdot 1 = 20x$, and here it is $-20x$, so the sign is negative.

$100x^2 - 20x + 1 = (10x - 1)^2$

The root of $100x^2$ is $10x$, so $(100x - 1)^2$ is wrong.

Use the difference pattern with $a = 2x$ and $b = \tfrac{1}{2}$. The first term is $(2x)^2 = 4x^2$, the middle is $-2\cdot 2x\cdot \tfrac{1}{2} = -2x$, and the last is $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.

$\left(2x - \tfrac{1}{2}\right)^2 = 4x^2 - 2x + \tfrac{1}{4}$

The doubling cancels the $\tfrac{1}{2}$, leaving a clean $-2x$.

For $49x^2 - 14x + 1$ the outer roots are $7x$ and $1$, and $2\cdot 7x\cdot 1 = 14x$ matches the middle term, with a minus sign.

$49x^2 - 14x + 1 = (7x - 1)^2$

The others fail: a perfect square needs a $+1$ last term, the middle $-7x$ is not $-14x$, and $49x^2 + 1$ has no middle term.

Use the difference pattern with first term $3a$ and last term $2b$.

$(3a - 2b)^2 = (3a)^2 - 2\cdot(3a)\cdot(2b) + (2b)^2 = 9a^2 - 12ab + 4b^2$

Each outer term is squared in full, and the middle term is $-12ab$.

The side length is the square root of the area, so factor the trinomial as a perfect square. The last term is $121 = 11^2$ and the middle is $2\cdot x\cdot 11 = 22x$.

$x^2 + 22x + 121 = (x + 11)^2$

So the side length is $x + 11$.

The area of a square is the side length squared, so compute $(x + 5)^2$.

$(x + 5)^2 = x^2 + 2\cdot x\cdot 5 + 5^2 = x^2 + 10x + 25$

The answer $4x + 20$ is the perimeter, not the area, and $x^2 + 25$ drops the middle term.

The last term of $(2x - b)^2$ is $b^2$, and here it is $25$, so $b^2 = 25$ gives $b = 5$. Check the middle: $-2\cdot 2x\cdot 5 = -20x$, which matches.

$(2x - 5)^2 = 4x^2 - 20x + 25$

So $b = 5$.