The Quadratic Formula: Complete the one solution equation

Let's solve the given equation:

$16a^2+20a+20=-5-20a$

First, let's organize the equation by moving terms and combining like terms:

$16a^2+20a+20=-5-20a \\ 16a^2+20a+20+5+20a =0\\ 16a^2+40a+25=0$

Now let's note that we can factor the expression on the left side using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+2\textcolor{red}{x}\textcolor{blue}{y}+\textcolor{blue}{y}^2$

We'll do this using the fact that:

$16=4^2\\ 25=5^2$

And using the law of exponents for powers applied to products in parentheses (in reverse):

$x^ny^n=(xy)^n$

Therefore, first we'll express the outer terms as a product of squared terms:

$16a^2+40a+25=0 \\ 4^2a^2+40a+5^2=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+40a+\textcolor{blue}{5}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2$

And the expression on the left side of the equation that we got in the last step:

$(\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0$

Note that the terms $(\textcolor{red}{4a})^2,\hspace{6pt}\textcolor{blue}{5}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2$

In other words - we ask if we can express the expression on the left side of the equation as:

$(\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ (\textcolor{red}{4a})^2+\underline{2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it holds that:

$2\cdot4a\cdot5=40a$

Therefore, we can express the expression on the left side of the equation as a perfect square binomial:

$(\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable and dividing both sides of the equation by the variable's coefficient:

$(4a+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ 4a+5=\pm0\\ 4a+5=0\\ 4a=-5\hspace{8pt}\text{/}:4\\ \boxed{a=-\frac{5}{4}}$

Let's summarize the solution of the equation:

$16a^2+20a+20=-5-20a \\ 16a^2+40a+25=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ 4a+5=0\\ \downarrow\\ \boxed{a=-\frac{5}{4}}$

Therefore the correct answer is answer D.

Let's solve the given equation:

$x^2+10x=-25$

First, let's arrange the equation by moving terms:

$x^2+10x=-25 \\ x^2+10x+25=0 \\$Now we notice that the expression on the left side can be factored using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We can do this using the fact that:

$25=5^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation we got in the last step:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a line):

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side as:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it is true that:

$2\cdot x\cdot5=10x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0$

From here we can take the square root of both sides of the equation (and don't forget there are two possibilities - positive and negative when taking the square root of an even power), then we'll easily solve by isolating the variable:

$(x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}$

Let's summarize the solution of the equation:

$x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$x^2-10x=-16$

First, let's arrange the equation by moving terms:

$x^2-10x=-16 \\ x^2-10x+16 =0$

Now we notice that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=16\\ m+n=-10\\$From the first requirement, namely - the multiplication, we notice that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same sign, according to multiplication rules, and now we'll remember that the possible factors of 16 are the number pairs 4 and 4, 2 and 8, or 16 and 1. Meeting the second requirement, along with the fact that the signs of the numbers we're looking for are identical will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=-2 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown in each:

$x-8=0\\ \boxed{x=8}$

or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$x^2-10x=-16 \\ x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=8,2}$

Therefore the correct answer is answer B.

Let's solve the given equation:

$y^2+4y+2=-2$

First, let's arrange the equation by moving terms:

$y^2+4y+2=-2 \\ y^2+4y+2+2=0 \\ y^2+4y+4=0$

Now we notice that the expression on the left side can be factored using the perfect square trinomial formula:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We can do this using the fact that:

$4=2^2$

Therefore, we'll represent the rightmost term as a squared term:

$y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+4y+\textcolor{blue}{2}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation we got in the last step:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2=0$

Notice that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{2}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2+\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}}+\textcolor{blue}{2}^2 =0$

And indeed it is true that:

$2\cdot y\cdot2=4y$

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

$\textcolor{red}{y}^2+2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}+\textcolor{blue}{2}^2=0 \\ \downarrow\\ (\textcolor{red}{y}+\textcolor{blue}{2})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(y+2)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y+2=\pm0\\ y+2=0\\ \boxed{y=-2}$

Let's summarize the solution of the equation:

$y^2+4y+2=-2 \\ y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}+\textcolor{blue}{2}^2=0 \\ \downarrow\\ (\textcolor{red}{y}+\textcolor{blue}{2})^2=0 \\ \downarrow\\ y+2=0\\ \downarrow\\ \boxed{y=-2}$

Therefore the correct answer is answer D.