# Solution by trinomial - Examples, Exercises and Solutions

More than once we have heard the teacher ask in class: "Who knows how to solve a quadratic equation without the formula?" We looked around to see who knew the answer, "Do you know what a trinomial is?" The teacher continued asking. We doubted and thought about what word this term could derive from and what a trinomial is. What does it really do? How does understanding about the trinomial benefit our mathematical knowledge? Does it expand the possibility of having greater mathematical efficiency? Or, in fact, might it be superfluous to include it in the ninth-grade curriculum?

In this article, we will try to answer these questions and even have fun with the properties of the trinomial that will help us quickly solve quadratic equations, to simplify fractions, to multiply and divide, to deal with fractions, even with the common denominator in fractions with variables in the numerator and in the denominator.

## Practice Solution by trinomial

### Exercise #1

$60-16y+y^2=-4$

### Step-by-Step Solution

Let's solve the given equation:

$60-16y+y^2=-4$First, let's arrange the equation by moving terms:

$60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0$Now, let's note that we can break down the expression on the left side using the short multiplication formula for the difference of two squares:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$This is done using the fact that:

$64=8^2$So let's present the outer term on the right as a square:

$y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0$Now let's examine again the short multiplication formula we 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 of the equation we got in the last step:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short multiplication formula we mentioned, the match to the short multiplication formula must also occur from the remaining term, meaning the middle term in the expression (highlighted with a bottom 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 it's possible to present the expression on the left side of the equation as:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0$And indeed it holds that:

$2\cdot y\cdot8=16y$So we can present the expression on the left side of the given equation as a difference of two squares:

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0$From here we can take out square roots for the two sides of the equation (and not forget that there are two possibilities - positive and negative when taking out square roots from an even order equation), we'll solve it easily by isolating the unknown in the side:

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

Let's summarize then the solution of the equation:

$60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}$

$y=8$

### Exercise #2

$4x^2=12x-9$

### Video Solution

$x=\frac{3}{2}$

### Exercise #3

$x^2+10x=-25$

### Video Solution

$x=-5$

### Exercise #4

What is the value of x?

$x^4-x^3=2x^2$

### Video Solution

$x=-1,2,0$

### Exercise #5

$x^2=6x-9$

### Video Solution

$x=3$

### Exercise #1

$x^2+144=24x$

### Video Solution

$x=12$

### Exercise #2

$x^2-10x=-16$

### Video Solution

$x=2,8$

### Exercise #3

Solve for y:

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

### Video Solution

$y=-2$

### Exercise #4

$x^3=x^2+2x$

### Video Solution

$x=0,-1,2$

### Exercise #5

Solve for x:

$x^2+32x=-256$

### Video Solution

$x=-16$

### Exercise #1

$3x^3-10x^2+7x=0$

### Video Solution

$x=0,1,\frac{7}{3}$

### Exercise #2

$x^3+x^2-12x=0$

### Video Solution

$x=0,3,-4$

### Exercise #3

$x^3-7x^2+6x=0$

### Video Solution

$x=0,1,6$