Quadratic Equation

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Lesson Plan ▼

Lesson Plan

The Standard Form

of a Quadratic Equation

$$\bbox[8pt]{\huge{ax^2+bx+c=0}}$$

$x$ represents an unknown

$a, b, c$ are the coefficients of the equation

$$\bbox[2pt]{\huge{a\neq 0}}$$

Coefficients of the Equation

$$\huge{ax^2+bx+c=0}$$

Coef.Name
$\Large{a}$Quadratic coefficient
$\Large{b}$Linear coefficient
$\Large{c}$Free term

Coefficients of the Equation

$$\Large{ax^2+bx+c=0}$$

$$\huge{2x^2+7x+3=0}$$

CoefficientValue
Quadratic coefficient$\Large{a=2}$
Linear coefficient$\Large{b=7}$
Free term$\Large{c=3}$

Determine the Coefficients

$$\Large{ax^2+bx+c=0}$$

$$\huge{x^2-5x-0.2=0}$$

CoefficientValue
Quadratic coefficient$\Large{a=1}$
Linear coefficient$\Large{b=-5}$
Free term$\Large{c=-0.2}$

Determine the Coefficients

$$\Large{2x-4+3x^2=0}$$

$$\huge{3x^2+2x-4=0}$$

CoefficientValue
Quadratic coefficient$\Large{a=3}$
Linear coefficient$\Large{b=2}$
Free term$\Large{c=-4}$

Types of Equations

TypeEquationDegree
Quadratic$ax^2+bx+c=0$ $\begin{array}{l} ax^2 \\ a\neq 0 \end{array} $
Linear$bx+c=0$ $\begin{array}{l} bx^1 \\ b\neq 0 \end{array} $

Determine the Type

EquationDegreeType
$x^2-3x+1=0$$2$Quadratic
$3x^2-7x+2=0$$2$Quadratic
$7x+12=0$$1$Linear
$2x-x^2+5=0$$2$Quadratic
$0\cdot x^2+x-8=0$$1$Linear

Root of an Equation

That value which, substituted for the unknown quantity in an equation, satisfies the equation

$$\Large{x^2+x-6=0}$$

If $\,x=2\,$ then $\,2^2+2-6=4+2-6=0$

If $\,x=-3\,$ then $\,(-3)^2-3-6=9-3-6=0$

So $2$ and $-3$ are roots

$$\Large{x\in \lbrace 2; -3\rbrace}$$

Check Roots

ValueEquationCheckRoot
$3 \rightarrow$$x^2-x-6=0$$9-3-6=0$ yes
$-1 \rightarrow$$x^2-x-2=0$$1+1-2\,\,$$= 0$yes
$2 \rightarrow$$x^2-2x+1=0$$2^2-4+1\,\,$$\neq 0$no
$1 \rightarrow$$x^2-3x+2=0$$1-3+2\,\,$$= 0$yes
$-2 \rightarrow$$x^2-2x-9=0$$4+4-9\,\,$$\neq 0$no

Examples of Equation

$ax^2+bx+c=0$$3x^2+5x-2=0$$a \neq 0$
$ax^2+bx=0$$2x^2+7x=0$ $\left\lbrace \begin{array}{l} c = 0 \\ a\neq 0 \end{array} \right .$
$ax^2-c=0$$4x^2-9=0$ $\left\lbrace \begin{array}{l} b = 0 \\ a\neq 0 \end{array} \right .$

$ax^2+bx+c=0$

$2x^2+5x=0$ $\left\lbrace \begin{array}{l} c = 0 \\ a\neq 0 \end{array} \right .$ $ax^2+bx=0$
$9-x^2=0$ $\left\lbrace \begin{array}{l} b = 0 \\ a\neq 0 \end{array} \right .$ $ax^2-c=0$
$3x+x^2=0$ $\left\lbrace \begin{array}{l} c = 0 \\ a\neq 0 \end{array} \right .$ $ax^2+bx=0$

$ax^2-c=0; \,\, (a\neq 0)$

$$\large{m^2-n^2=(m-n)(m+n)}$$

$x^2-2^2=0;$$(x-2)(x+2)=0;$$\left[\begin{array}{l} x-2=0\\ x+2=0 \end{array}\right.$
$x^2-25=0;$$(x-5)(x+5)=0;$ $\left[\begin{array}{l} x=5\\ x=-5 \end{array}\right. $ ☑
$4x^2-1=0;$$(2x-1)(2x+1)=0;$ $\left[\begin{array}{l} 2x=1\\ 2x=-1 \end{array}\right. $

Factoring the Quadratic Equation

$$\bbox[1em]{\large{ax^2-c=0;\quad a\neq 0}}$$

$x^2-16=0;\quad$ $(x-4)(x+4)=0\quad$

$x\in \lbrace 4; -4\rbrace ;$ $\quad\left[\begin{array}{l} x-4=0\\ x+4=0 \end{array}\right. ;\quad $ $\left[\begin{array}{l} x=4\\ x=-4 \end{array}\right. $

Factoring the Quadratic Equation

$$\bbox[1em]{\large{ax^2-c=0;\quad a\neq 0}}$$

$x^2-2=0;\quad$ $(x-\sqrt{2})(x+\sqrt{2})=0\quad$

$x\in \lbrace\sqrt{2}; -\sqrt{2}\rbrace ;$ $\quad\left[\begin{array}{l} x-\sqrt{2}=0\\ x+\sqrt{2}=0 \end{array}\right. ;\quad $ $\left[\begin{array}{l} x=\sqrt{2}\\ x=-\sqrt{2} \end{array}\right. $

Factoring the Quadratic Equation

$$\bbox[1em]{\large{ax^2-c=0;\quad a\neq 0}}$$

$9x^2-64=0;\quad$ $(3x-8)(3x+8)=0\quad$

$x\in \left\lbrace 2\dfrac{2}{3}; -2\dfrac{2}{3}\right\rbrace ;$ $\quad\left[\begin{array}{l} 3x-8=0\\ 3x+8=0 \end{array}\right. ;\quad $ $\left[\begin{array}{l} x=\dfrac{8}{3}\\ x=-\dfrac{8}{3} \end{array}\right. $

Factoring the Quadratic Equation

$$\bbox[1em]{\large{ax^2-c=0;\quad a\neq 0}}$$

$2x^2-18=0;\quad$ $(x-3)(x+3)=0\quad$

$2(x^2-9)=0;$ $\quad\left[\begin{array}{l} x-3=0\\ x+3=0 \end{array}\right. ;\quad $ $\left[\begin{array}{l} x=3\\ x=-3 \end{array}\right. $

$x\in \lbrace 3; -3\rbrace ;$

Solving the Quadratic Equation

$\large{ax^2-c=0\,\,}$ $\large{\Rightarrow \,\, ax^2=c\,\,}$$\large{\Rightarrow \,\, x^2=\dfrac{c}{a}\,\,}$

$\large{\Rightarrow \,\, x=\pm\sqrt{\dfrac{c}{a}};\,\, \left(a\neq 0; \dfrac{c}{a}\geqslant 0\right)}$

$2x^2-32=0\,$$\Rightarrow\, 2x^2=32\,$$\Rightarrow\, x^2=\dfrac{32}{2}=16$

$x=\pm\sqrt{16};\,\,$$x\in \lbrace 4; -4\rbrace$

Solving the Quadratic Equation

$\large{ax^2-c=0\,\,}$ $\large{\Rightarrow \,\, ax^2=c\,\,}$ $\large{\Rightarrow \,\, x^2=\dfrac{c}{a}\,\,}$

$\large{\Rightarrow \,\, x=\pm\sqrt{\dfrac{c}{a}};\,\, \left(a\neq 0; \dfrac{c}{a}\geqslant 0\right)}$

$2x^2-50=0\,$$\Rightarrow\, 2x^2=50\,$$\Rightarrow\, x^2=\dfrac{50}{2}=25$

$x=\pm\sqrt{25};\,\,$$x\in \lbrace 5; -5\rbrace$

Solving the Quadratic Equation

$$ax^2-c=0\,\,\Rightarrow x=\pm\sqrt{\dfrac{c}{a}};\,\, \left(a\neq 0; \dfrac{c}{a}\geqslant 0\right)$$

$x^2-100=0;$$x=\pm\sqrt{100};$$x\in \lbrace 10; -10\rbrace$
$5x^2+10=0;$$x^2=-2;$$\varnothing$
$x^2-1=0;$$x=\pm\sqrt{1};$$x\in \lbrace 1; -1\rbrace$
$6x^2-12=0;$$x=\pm\sqrt{2};$$x\in \lbrace \sqrt{2}; -\sqrt{2}\rbrace$
$9x^2-1=0;$$x=\pm\sqrt{\dfrac{1}{9}};$$x\in \left\lbrace \dfrac{1}{3}; -\dfrac{1}{3}\right\rbrace$

$ax^2+bx=0; \,\, (a\neq 0)$

$ax^2+bx=0\, \Rightarrow x(ax+b)=0\,\,$$\Rightarrow \left[\begin{array}{l} x=0\\ x=-\dfrac{b}{a} \end{array}\right.$

$x^2-4x=0;$$x(x-4)=0;$$\left[\begin{array}{l} x=0\\ x=4 \end{array}\right.$ ☑
$x^2+x=0;$$x(x+1)=0;$ $\left[\begin{array}{l} x=0\\ x=-1 \end{array}\right. $

Factoring the Quadratic Equation

$$ax^2+bx=0\,\, \Rightarrow ax\left(x+\dfrac{b}{a}\right)=0$$

$2x^2-4x=0;$$2x(x-2)=0;$$x\in \lbrace 0; 2\rbrace$
$3x^2+6x=0;$$3x(x+2)=0;$$x\in \lbrace 0; -2\rbrace$
$2x^2-3x=0;$$2x\left(x-\dfrac{3}{2}\right)=0;$$x\in \left\lbrace 0; 1.5\right\rbrace$
$5x^2+x=0;$$5x\left(x-\dfrac{1}{5}\right)=0;$$x\in \left\lbrace 0; \dfrac{1}{5}\right\rbrace$

$ax^2+bx+c=0; \,\, (a\neq 0)$

$$\large{1) \quad D=b^2-4ac}$$

  • if $D < 0$ then there are no real roots
  • if $D \geqslant 0$ then the equation has roots

$$\large{2) \quad x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}}$$

Solving the Quadratic Equation

$\begin{array}{c} ax^2+bx+c=0;\\ D=b^2-4ac; \end{array} $ $\left\lbrace\begin{array}{l} a\neq 0\\ D\geqslant 0 \end{array}\right. ; $ $x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$

$\bbox[2pt]{\large{x^2-3x+2=0}}$

$\begin{array}{l} a=1;\\ b=-3;\\ c=2; \end{array}\, $ $\begin{array}{l} D=(-3)^2-4\cdot 1 \cdot 2;\\ D=9-8=1;\\ D\geqslant 0;\, \sqrt{D}=1; \end{array}\,\, $ $\begin{array}{l} x_{1}=\dfrac{3+1}{2\cdot 1}\\ x_{2}=\dfrac{3-1}{2\cdot 1}\\ \end{array} $

$x\in \left\lbrace 2; 1\right\rbrace$

Solving the Quadratic Equation

$\begin{array}{c} ax^2+bx+c=0;\\ D=b^2-4ac; \end{array} $ $\left\lbrace\begin{array}{l} a\neq 0\\ D\geqslant 0 \end{array}\right. ; $ $x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$

$\bbox[2pt]{\large{x^2-x-6=0}}$ ☑

$\begin{array}{l} a=1;\\ b=-1;\\ c=-6; \end{array}\, $ $\begin{array}{l} D=(-1)^2-4\cdot (-6);\\ D=1+24=25;\\ D\geqslant 0;\, \sqrt{25}=5; \end{array}\,\, $ $\begin{array}{l} x_{1}=\dfrac{1+5}{2\cdot 1}\\ x_{2}=\dfrac{1-5}{2\cdot 1}\\ \end{array} $

$x\in \left\lbrace 3; -2\right\rbrace$

Solving the Quadratic Equation

$\begin{array}{c} ax^2+bx+c=0;\\ D=b^2-4ac; \end{array} $ $\left\lbrace\begin{array}{l} a\neq 0\\ D\geqslant 0 \end{array}\right. ; $ $x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$

$\bbox[2pt]{\large{x^2-3x+2=0}}$

$\begin{array}{l} a=1;\\ b=-3;\\ c=2; \end{array}\, $ $\begin{array}{l} D=(-3)^2-4\cdot 1 \cdot 2;\\ D=9-8=1;\\ D\geqslant 0;\, \sqrt{D}=1; \end{array}\,\, $ $\begin{array}{l} x_{1}=\dfrac{3+1}{2\cdot 1}\\ x_{2}=\dfrac{3-1}{2\cdot 1}\\ \end{array} $

$x\in \left\lbrace 2; 1\right\rbrace$

Solving the Quadratic Equation

$\begin{array}{c} ax^2+bx+c=0;\\ D=b^2-4ac; \end{array} $ $\left\lbrace\begin{array}{l} a\neq 0\\ D\geqslant 0 \end{array}\right. ; $ $x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$

$\bbox[2pt]{\large{x^2+2x+6=0}}$

$\begin{array}{l} a=1;\\ b=2;\\ c=6; \end{array}\, $ $\begin{array}{l} D=2^2-4\cdot 1 \cdot 6;\\ D=4-24=-20;\\ D< 0; \end{array}\,\, $

$\varnothing$

Solving the Quadratic Equation

$\begin{array}{c} ax^2+bx+c=0;\\ D=b^2-4ac; \end{array} $ $\left\lbrace\begin{array}{l} a\neq 0\\ D\geqslant 0 \end{array}\right. ; $ $x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$

$x^2+5x+6=0$$D=25-24$$x_{1,2}=\dfrac{-5\pm 1}{2}$
$x^2+5x+4=0$$D=25-16$$x_{1,2}=\dfrac{-5\pm 3}{2}$
$x^2-3x-10=0$$D=9+40$$x_{1,2}=\dfrac{3\pm 7}{2}$

Vieta's Formulas

If $x_1$ and $x_2$ are the roots of the quadratic equation

$$\large{x^2+bx+c=0;\quad (a=1)}$$

then

$$\large{ \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & c \\ x_{1} + x_{2} & = & -b \end{array} \right .} $$

Vieta's Formulas

$\bbox[12pt]{ \begin{array}{c} x^2+bx+c=0 \\ x\in \left\lbrace x_1; x_2 \right\rbrace \end{array}; \quad \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & c \\ x_{1} + x_{2} & = & -b \end{array} \right .} $

$\large{x^2-3x+2=0}$

$\begin{array}{ll} b=-3; & x_1=2; \\ c=2; & x_2=1; \end{array}\,\,\, $ $\left\lbrace \begin{array}{lcrcr} 1\cdot 2 & = & 2 & = &c\\ 1 + 2 & = & 3 & = &-b \end{array} \right . $

Vieta's Formulas

$\bbox[12pt]{ \begin{array}{c} x^2+bx+c=0 \\ x\in \left\lbrace x_1; x_2 \right\rbrace \end{array}; \quad \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & c \\ x_{1} + x_{2} & = & -b \end{array} \right .} $

$\large{x^2-x-6=0}$

$\begin{array}{ll} b=-1; & x_1=3; \\ c=-6; & x_2=-2; \end{array}\,\,\, $ $\left\lbrace \begin{array}{lcr} 3\cdot (-2) & = & -6\\ 3 + (-2) & = & 1 \end{array} \right . $

Vieta's Formulas

$\bbox[12pt]{ \begin{array}{c} x^2+bx+c=0 \\ x\in \left\lbrace x_1; x_2 \right\rbrace \end{array}; \quad \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & c \\ x_{1} + x_{2} & = & -b \end{array} \right .} $

$\large{x^2-4x-5=0}$

$\begin{array}{ll} b=-4; & x_1=5; \\ c=-5; & x_2=-1; \end{array}\,\,\, $ $\left\lbrace \begin{array}{lcr} 5\cdot (-1) & = & -5\\ 5 + (-1) & = & 4 \end{array} \right . $

Vieta's Formulas

If $x_1$ and $x_2$ are the roots of the quadratic equation

$$\large{ax^2+bx+c=0;\quad (a\neq 0)}$$

then

$$\large{ \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & \dfrac{c}{a} \\ x_{1} + x_{2} & = & -\dfrac{b}{a} \end{array} \right .} $$

Vieta's Formulas

$\begin{array}{c} ax^2+bx+c=0 \\ x\in \left\lbrace x_1; x_2 \right\rbrace ,\, (a\neq 0) \end{array}; \quad \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & \dfrac{c}{a} \\ x_{1} + x_{2} & = & -\dfrac{b}{a} \end{array} \right . $

$\large{2x^2+4x-6=0}$

$\begin{array}{ll} a=2; & x_1=1; \\ b=4; & x_2=-3; \\ c=-6;& \end{array}$ $\left\lbrace \begin{array}{ll} 1\cdot (-3) &= -3 = \dfrac{-6}{2} \\ 1 + (-3) &= -2 = -\dfrac{4}{2} \end{array} \right . $

Vieta's Formulas

$\begin{array}{c} ax^2+bx+c=0 \\ x\in \left\lbrace x_1; x_2 \right\rbrace ,\, (a\neq 0) \end{array}; \quad \left\lbrace \begin{array}{lcr} x_{1}\cdot x_{2} & = & \dfrac{c}{a} \\ x_{1} + x_{2} & = & -\dfrac{b}{a} \end{array} \right . $

$\large{3x^2+5x-2=0}$

$\begin{array}{ll} a=3; & x_1=\dfrac{1}{3}; \\ b=5; & x_2=-2; \\ c=-2;& \end{array}$ $\left\lbrace \begin{array}{ll} \dfrac{1}{3}\cdot (-2) &= \dfrac{-2}{3} \\ \dfrac{1}{3}+(-2) &=-\dfrac{5}{3} \end{array} \right . $

How to Solve Quadratic Equations

$$ax^2+bx+c=0;\,\, (a\neq 0)$$

1) Find the solution

$$D=b^2-4ac\geqslant 0;\,\, x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$$

2) Use Vieta's formulas to check the solution

$$\left\lbrace \begin{array}{ccr} x_1 \cdot x_2 & = & \dfrac{c}{a} \\ x_1 + x_2 & = & -\dfrac{b}{a} \end{array}\right . $$

Solve the equation, check the roots

$$\Large{x^2-5x+6=0}$$

$ \begin{array}{c} ax^2+bx+c=0\\ D=b^2-4ac \\ x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a} \end{array} $ $\left\lbrace \begin{array}{l} x_1 \cdot x_2 = \dfrac{c}{a} \\ x_1 + x_2 = -\dfrac{b}{a} \end{array}\right . $
$ \begin{array}{c} D=25-24=1\\ x_{1,2}=\dfrac{5\pm 1}{2}\\ x\in \left\lbrace 2; 3 \right\rbrace \end{array} $ $\left\lbrace \begin{array}{l} 2 \cdot 3 = \dfrac{6}{1} \\ 2 + 3 = -\dfrac{(-5)}{1} \end{array}\right . $

Solve the equation, check the roots

$$\Large{x^2+x-2=0}$$

$ \begin{array}{c} ax^2+bx+c=0\\ D=b^2-4ac \\ x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a} \end{array} $ $\left\lbrace \begin{array}{l} x_1 \cdot x_2 = \dfrac{c}{a} \\ x_1 + x_2 = -\dfrac{b}{a} \end{array}\right . $
$ \begin{array}{c} D=1+8=9\\ x_{1,2}=\dfrac{-1\pm 3}{2}\\ x\in \left\lbrace 1; -2 \right\rbrace \end{array} $ $\left\lbrace \begin{array}{l} 1 \cdot (-2) = -2 \\ 1 - 2 = -1 \end{array}\right . $

Solve the equation, check the roots

$$\Large{2x^2+x-1=0}$$

$ \begin{array}{c} ax^2+bx+c=0\\ D=b^2-4ac \\ x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a} \end{array} $ $\left\lbrace \begin{array}{l} x_1 \cdot x_2 = \dfrac{c}{a} \\ x_1 + x_2 = -\dfrac{b}{a} \end{array}\right . $
$ \begin{array}{c} D=1+8=9\\ x_{1,2}=\dfrac{-1\pm 3}{2}\\ x\in \left\lbrace 0.5; -1 \right\rbrace \end{array} $ $\left\lbrace \begin{array}{l} \dfrac{1}{2} \cdot (-1) = -\dfrac{1}{2} \\ \dfrac{1}{2} - 1 = -\dfrac{1}{2} \end{array}\right . $

Factoring the Quadratic Equation

$$\bbox[10pt]{\large{ax^2+bx+c=a\left( x-x_1\right)\left( x-x_2\right)}}$$

where $x_1$ and $x_2$ are the roots of the equation

$$ax^2+bx+c=0;\,\, (a\neq 0)$$

Factoring the Quadratic Equation

$$\large{ax^2+bx+c=a\left( x-x_1\right)\left( x-x_2\right)}$$

EquationResult
$\begin{array}{c} x^2-3x+2=0;\\ x_1= 2;\, x_2=1; \end{array}$ $\begin{array}{l} (x-2)(x-1)=0\\ \qquad x_1\qquad x_2 \end{array}$
$\begin{array}{c} x^2-x-6=0;\\ x_1 = 3; x_2 = -2; \end{array}$ $\begin{array}{l} (x-3)(x-(-2))=0\\ (x-3)(x+2)=0 \end{array}$
$\begin{array}{l} x^2+5x+6=0;\\ x_1 = -3; x_2 = -2; \end{array}$ $\begin{array}{l} (x-(-3))(x-(-2))=0\\ (x+3)(x+2)=0 \end{array}$

Factoring the Quadratic Equation

$$\large{ax^2+bx+c=a\left( x-x_1\right)\left( x-x_2\right)}$$

EquationResult
$\begin{array}{l} x^2-5x+6=0;\\ x\in \left\lbrace 2; 3\right\rbrace \end{array}$$(x-2)(x-3)=0$
$\begin{array}{l} x^2-x-2=0;\\ x\in \left\lbrace 2; -1\right\rbrace \end{array}$$(x-2)(x+1)=0$
$\begin{array}{l} x^2-2x-8=0;\\ x\in \left\lbrace -2; 4\right\rbrace \end{array}$$(x+2)(x-4)=0$

Factoring the Quadratic Equation

$$\large{ax^2+bx+c=a\left( x-x_1\right)\left( x-x_2\right)}$$

EquationResult
$\begin{array}{l} 2x^2-2x-4=0;\\ x\in \left\lbrace 2; -1 \right\rbrace \end{array}$ $\begin{array}{l} 2(x-2)(x+1)=0\\ a\qquad x_1\qquad x_2 \end{array}$
$\begin{array}{l} 3x^2+3x-6=0;\\ x\in \left\lbrace -2; 1 \right\rbrace \end{array}$$3(x+2)(x-1)=0$

Factoring the Quadratic Equation

$$\large{ax^2+bx+c=a\left( x-x_1\right)\left( x-x_2\right)}$$

EquationResult
$\begin{array}{l} 2x^2+7x+3=0;\\ x\in \left\lbrace 3; \dfrac{1}{2} \right\rbrace \end{array}$$2(x-3)\left(x-\dfrac{1}{2}\right)=0$
$ \begin{array}{l} 6x^2-5x+1=0;\\ x\in \left\lbrace \dfrac{1}{3}; \dfrac{1}{2}\right\rbrace \end{array}$$6\left(x-\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0$

Factoring the Quadratic Equation

$$\bbox[5pt]{\large{ax^2+bx+c=a\left( x-x_1\right)\left( x-x_2\right)}}$$

$\toggle{x^2-x-20=0;\,x\in \left\lbrace 5; -4\right\rbrace}{x^2-x-20=(x-5)(x+4)}\endtoggle$

$\toggle{2x^2-5x-3=0;\,x\in \left\lbrace 3; -\dfrac{1}{2}\right\rbrace}{2x^2-5x-3=2(x-3)\left(x+\dfrac{1}{2}\right)}\endtoggle$

$\toggle{3x^2-7x+3=0;\,x\in \left\lbrace 2; \dfrac{1}{3}\right\rbrace}{3x^2-7x+3=3(x-2)\left(x-\dfrac{1}{3}\right)}\endtoggle$

$\toggle{10x^2-7x+1=0;\,x\in \left\lbrace \dfrac{1}{5}; \dfrac{1}{2}\right\rbrace}{10x^2-7x+1=10\left(x-\dfrac{1}{5}\right)\left(x-\dfrac{1}{2}\right)}\endtoggle$

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