#### $ax^2+bx+c$

Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the style of a trinomial, the exercise will be called "trinomial".

Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the style of a trinomial, the exercise will be called "trinomial".

$(x+solution \space one)(x+solution\space two)$

or with subtractions, depending on the solutions.

Question 1

How many solutions does the equation have?

\( x^4+12x^3+36x^2=0 \)

Question 2

\( x^2+6x+9=0 \)

What is the value of X?

Question 3

\( x^2-3x+2=0 \)

What is the value of X?

Question 4

\( x^2-12x+36=0 \)

What is the value of X?

Question 5

\( -x^2+13x-14=0 \)

What is the value of X?

How many solutions does the equation have?

$x^4+12x^3+36x^2=0$

Let's solve the given equation:

$x^4+12x^3+36x^2=0$We note that it is possible __to factor the expression__ which is in the left side of the given equation, this is done by __taking out the common factor__ $x^2$ which is the __greatest common factor of the numbers and letters__ in the expression:

$x^4+12x^3+36x^2=0 \\
\downarrow\\
x^2(x^2+12x+36) =0$**We continue** and refer to the expression on the left side of the equation that we got in the last step which contains algebraic expressions and on the right side the number 0, therefore, **since the only way to get the result 0 from a product is to multiply by 0**, __at least__ one of the expressions in the product on the left side, must be equal to zero,

Meaning:

$x^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=0}$

__Or:__

$x^2+12x+36=0$** We continue** and in order to find the

__Note that the leading coefficient is 1__, so we can (try) to solve it using __the quick trinomial formula,__

However, we can factor, in this case, also using __the short multiplication formula for a binomial squared:__

$(\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$**The motivation** for trying factoring in this approach is the fact that we can identify in the left side expression in the equation we got in the last step, that the two terms which are **in the extreme positions** (meaning the term in the first position - it is the squared term and the term in the zero position - it is the free number in the expression) can be presented (simply) as a squared term:

$x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0$

Equating the expression on the left side **in the equation:**

$\downarrow\\ x^2+12x+6^2$

To the expression on the right side **in the short multiplication formula above:**

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

The conclusion from this is that **what remains to check is whether the middle term in the expression in the equation** matches the middle term in the short multiplication formula above, meaning - after identifying $a$$b$which are both in the first position in the short multiplication formula above in which $a$and $b$we check if the middle term in the expression in the left side of the equation **can be presented** as $2\cdot a \cdot b$we check if so, we start by presenting the equality of the short multiplication formula to the given expression:

$\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 \Leftrightarrow \textcolor{red}{x}^2+\underline{12x}+\textcolor{blue}{6}^2$__And__ ** indeed it holds** that:

$2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}=12x$Meaning **the middle term in the expression in the equation indeed matches the form of the middle term in the short multiplication formula** (highlighted with a line below), mathematically:

$x^2+\underline{12x}+6^2=0 \\
\textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0\\
\downarrow\\
(\textcolor{red}{x}+\textcolor{blue}{6})^2=0$We can now remember that **a real root can be calculated only for a positive number or for the number zero** (since __it is not possible to get a negative number from squaring a real number itself__), and therefore for an equation there are two real solutions (or one solution) only if:

**Next** we note that if: $(x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\
x+6=0\\
\boxed{x=-6}$then the __only__ solution to the equation is:

$x^4+12x^3+36x^2=0 \\
\downarrow\\
x^2(x^2+12x+36) =0 \\
\downarrow\\
x^2=0\rightarrow\boxed{x=0} \\
x^2+12x+36=0\\
x^2+2\cdot2\cdot6+6^2=0\\
\rightarrow(x+6)^2=0\rightarrow\boxed{x=-6}\\
\downarrow\\
\boxed{x=0,-6}$Therefore, ** we can summarize what was explained in the rule**:

In the quadratic equation:

$ax^2+bx+c =0$in which the coefficients are substituted and the discriminant is calculated:

$a,b$__If it holds:__

** a.**$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:

__There is no (real) solution__ to the equation.

** b.**$\Delta$:

__There exists a single (real) solution__ to the equation.

** c.**$x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac$:

__There exist two (real) solutions__ to the equation.

** Now let's return to the given equation** and extract from it the

$\Delta\geq0$We continue and __calculate $\Delta=0$:__

$x=-\frac{b}{2a}$Therefore for the quadratic equation that we solved, ** one (real) solution**,

and in combination with the solution $ax^2+bx+c =0$(the additional solution we found for the given equation which is indicated in the first step after factoring using the common factor),

**Therefore we get that for the given equation:**

$\Delta=b^2-4ac$

__two real solutions.__

Two solutions

$x^2+6x+9=0$

What is the value of X?

3-

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

What is the value of X?

$1,2$

$x^2-12x+36=0$

What is the value of X?

6

$-x^2+13x-14=0$

What is the value of X?

$x_1=14,x_2=-1$

Question 1

\( 2x^2+4x-6=0 \)

What is the value of X?

Question 2

\( x^2-1=0 \)

Question 3

\( x^2-5x-50=0 \)

Question 4

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

Question 5

\( x^2+x-2=0 \)

$2x^2+4x-6=0$

What is the value of X?

$x_1=1,x_2=-3$

$x^2-1=0$

$x=\pm1$

$x^2-5x-50=0$

$x=10,x=-5$

$x^2-8x+16=0$

$x=4$

$x^2+x-2=0$

$(x-1)(x+2)=0$

Question 1

\( x^2+9x+20=0 \)

Question 2

\( x^2-2x-3=0 \)

Question 3

\( x^2+6x+9=0 \)

Question 4

\( x^2-7x+12=0 \)

Question 5

\( x^2-19x+60=0 \)

$x^2+9x+20=0$

$x=-4,x=-5$

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

$x=3,x=-1$

$x^2+6x+9=0$

$x=-3$

$x^2-7x+12=0$

$x=3,x=4$

$x^2-19x+60=0$

$x=15,x=4$