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Formula
Number of solution
Relationship between roots and coefficients
$a ^{ 2 } = 12$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = - 2 \sqrt{ 3 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 12 }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ a } = \pm \sqrt{ \color{#FF6800}{ 12 } }$
$\color{#FF6800}{ a } = \pm \sqrt{ \color{#FF6800}{ 12 } }$
 Solve a solution to $a$
$\color{#FF6800}{ a } = \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } }$
$\color{#FF6800}{ a } = \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ a } = \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ a } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = - 2 \sqrt{ 3 } \end{array}$
$a ^ { 2 } = \color{#FF6800}{ 12 }$
 Move the expression to the left side and change the symbol 
$a ^ { 2 } - 12 = 0$
$a = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
 0 has no sign 
$a = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
$a = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
 The power of 0 is 0 
$a = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 0 - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 48 } } { 2 \times 1 } }$
$a = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 48 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$a = \dfrac { 0 \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { 2 \times 1 }$
$a = \dfrac { 0 \pm 4 \sqrt{ 3 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$a = \dfrac { 0 \pm 4 \sqrt{ 3 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 0 \pm 4 \sqrt{ 3 } } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 0 + 4 \sqrt{ 3 } } { 2 } } \\ \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } } \end{array}$
$\begin{array} {l} a = \dfrac { \color{#FF6800}{ 0 } + 4 \sqrt{ 3 } } { 2 } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
 0 does not change when you add or subtract 
$\begin{array} {l} a = \dfrac { 4 \sqrt{ 3 } } { 2 } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = \color{#FF6800}{ \dfrac { 4 \sqrt{ 3 } } { 2 } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} a = \color{#FF6800}{ \dfrac { 2 \sqrt{ 3 } } { 1 } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = \dfrac { 2 \sqrt{ 3 } } { \color{#FF6800}{ 1 } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} a = \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \dfrac { \color{#FF6800}{ 0 } - 4 \sqrt{ 3 } } { 2 } \end{array}$
 0 does not change when you add or subtract 
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \dfrac { - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \color{#FF6800}{ \dfrac { - 4 \sqrt{ 3 } } { 2 } } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } } { 1 } } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \dfrac { - 2 \sqrt{ 3 } } { \color{#FF6800}{ 1 } } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \end{array}$
 2 real roots 
Find the number of solutions
$a ^ { 2 } = \color{#FF6800}{ 12 }$
 Move the expression to the left side and change the symbol 
$a ^ { 2 } - 12 = 0$
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 12 \right )$
 The power of 0 is 0 
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 12 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 12 \right )$
 0 does not change when you add or subtract 
$D = - 4 \times 1 \times \left ( - 12 \right )$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 12 \right )$
 Multiplying any number by 1 does not change the value 
$D = - 4 \times \left ( - 12 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$
 Multiply $- 4$ and $- 12$
$D = \color{#FF6800}{ 48 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 48 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = 0 , \alpha \beta = - 12$
Find the sum and product of the two roots of the quadratic equation
$a ^ { 2 } = \color{#FF6800}{ 12 }$
 Move the expression to the left side and change the symbol 
$a ^ { 2 } - 12 = 0$
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 12 } { 1 } }$
$\alpha + \beta = - \dfrac { 0 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 12 } { 1 }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 1 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 1 }$
 0 has no sign 
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 1 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { - 12 } { \color{#FF6800}{ 1 } }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
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