Symbol

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Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = x ^ { 2 } + 2 x - 4$
$y = 0$
$x$Intercept
$\left ( - \sqrt{ 5 } - 1 , 0 \right )$, $\left ( - 1 + \sqrt{ 5 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 4 \right )$
Minimum
$\left ( - 1 , - 5 \right )$
Standard form
$y = \left ( x + 1 \right ) ^ { 2 } - 5$
$x ^{ 2 } +2x-4 = 0$
$\begin{array} {l} x = - 1 + \sqrt{ 5 } \\ x = - 1 - \sqrt{ 5 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + 1 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x + 1 \right ) ^ { 2 } = 4 + \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
 Calculate power 
$\left ( x + 1 \right ) ^ { 2 } = 4 + \color{#FF6800}{ 1 }$
$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
 Add $4$ and $1$
$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 5 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 5 }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ 5 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ 5 } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 5 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 5 } } \end{array}$
$\begin{array} {l} x = - 1 + \sqrt{ 5 } \\ x = - 1 - \sqrt{ 5 } \end{array}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ 2 ^ { 2 } - 4 \times 1 \times \left ( - 4 \right ) } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ 20 } } { 2 \times 1 } }$
$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ 20 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$x = \dfrac { - 2 \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 5 } } } { 2 \times 1 }$
$x = \dfrac { - 2 \pm 2 \sqrt{ 5 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { - 2 \pm 2 \sqrt{ 5 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm 2 \sqrt{ 5 } } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 + 2 \sqrt{ 5 } } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 - 2 \sqrt{ 5 } } { 2 } } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 2 + 2 \sqrt{ 5 } } { 2 } } \\ x = \dfrac { - 2 - 2 \sqrt{ 5 } } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 1 + \sqrt{ 5 } } { 1 } } \\ x = \dfrac { - 2 - 2 \sqrt{ 5 } } { 2 } \end{array}$
$\begin{array} {l} x = \dfrac { - 1 + \sqrt{ 5 } } { \color{#FF6800}{ 1 } } \\ x = \dfrac { - 2 - 2 \sqrt{ 5 } } { 2 } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 5 } } \\ x = \dfrac { - 2 - 2 \sqrt{ 5 } } { 2 } \end{array}$
$\begin{array} {l} x = - 1 + \sqrt{ 5 } \\ x = \color{#FF6800}{ \dfrac { - 2 - 2 \sqrt{ 5 } } { 2 } } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = - 1 + \sqrt{ 5 } \\ x = \color{#FF6800}{ \dfrac { - 1 - \sqrt{ 5 } } { 1 } } \end{array}$
$\begin{array} {l} x = - 1 + \sqrt{ 5 } \\ x = \dfrac { - 1 - \sqrt{ 5 } } { \color{#FF6800}{ 1 } } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} x = - 1 + \sqrt{ 5 } \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 5 } } \end{array}$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right )$
$D = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 4 \right )$
 Calculate power 
$D = \color{#FF6800}{ 4 } - 4 \times 1 \times \left ( - 4 \right )$
$D = 4 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 4 \right )$
 Multiplying any number by 1 does not change the value 
$D = 4 - 4 \times \left ( - 4 \right )$
$D = 4 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right )$
 Multiply $- 4$ and $- 4$
$D = 4 + \color{#FF6800}{ 16 }$
$D = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
 Add $4$ and $16$
$D = \color{#FF6800}{ 20 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 20 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = - 2 , \alpha \beta = - 4$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 4 } { 1 } }$
$\alpha + \beta = - \dfrac { 2 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 4 } { 1 }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = - \color{#FF6800}{ 2 } , \alpha \beta = \dfrac { - 4 } { 1 }$
$\alpha + \beta = - 2 , \alpha \beta = \dfrac { - 4 } { \color{#FF6800}{ 1 } }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = - 2 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
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