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Solve the quadratic equation

Answer

Number of solution

Answer

Relationship between roots and coefficients

Answer

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$y = x ^ { 2 } + 2 x + 6$

$y = 0$

$y$-intercept

$\left ( 0 , 6 \right )$

Minimum

$\left ( - 1 , 5 \right )$

Standard form

$y = \left ( x + 1 \right ) ^ { 2 } + 5$

$x ^{ 2 } +2x+6 = 0$

$\begin{array} {l} x = - 1 + \sqrt{ 5 } i \\ x = - 1 - \sqrt{ 5 } i \end{array}$

Solve quadratic equations using the square root

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$

$ $ Convert the quadratic expression on the left side to a perfect square format $ $

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$

$\left ( x + 1 \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = 0$

$ $ Move the constant to the right side and change the sign $ $

$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$

$\left ( x + 1 \right ) ^ { 2 } = - 6 + \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$

$ $ Calculate power $ $

$\left ( x + 1 \right ) ^ { 2 } = - 6 + \color{#FF6800}{ 1 }$

$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$

$ $ Add $ - 6 $ and $ 1$

$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$

$ $ Solve quadratic equations using the square root $ $

$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 5 } }$

$ $ Solve a solution to $ x$

$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ i } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ i } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ i } \end{array}$

$\begin{array} {l} x = - 1 + \sqrt{ 5 } i \\ x = - 1 - \sqrt{ 5 } i \end{array}$

Calculate using the quodratic formula$($Imaginary root solution$)

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$

$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ 2 ^ { 2 } - 4 \times 1 \times 6 } } { 2 \times 1 } }$

$ $ Organize the expression $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ - 20 } } { 2 \times 1 } }$

$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 20 } } } { 2 \times 1 }$

$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $

$x = \dfrac { - 2 \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ i } } { 2 \times 1 }$

$x = \dfrac { - 2 \pm 2 \sqrt{ 5 } i } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$

$ $ Multiplying any number by 1 does not change the value $ $

$x = \dfrac { - 2 \pm 2 \sqrt{ 5 } i } { \color{#FF6800}{ 2 } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm 2 \sqrt{ 5 } i } { 2 } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 + 2 \sqrt{ 5 } i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 - 2 \sqrt{ 5 } i } { 2 } } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 2 + 2 \sqrt{ 5 } i } { 2 } } \\ x = \dfrac { - 2 - 2 \sqrt{ 5 } i } { 2 } \end{array}$

$ $ Reduce the fraction $ $

$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ i } \\ x = \dfrac { - 2 - 2 \sqrt{ 5 } i } { 2 } \end{array}$

$\begin{array} {l} x = - 1 + \sqrt{ 5 } i \\ x = \color{#FF6800}{ \dfrac { - 2 - 2 \sqrt{ 5 } i } { 2 } } \end{array}$

$ $ Reduce the fraction $ $

$\begin{array} {l} x = - 1 + \sqrt{ 5 } i \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 5 } } \color{#FF6800}{ i } \end{array}$

$ $ Do not have the solution $ $

Calculate using the quadratic formula

$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 6 } } { 2 \times 1 }$

$ $ Calculate power $ $

$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ 4 } - 4 \times 1 \times 6 } } { 2 \times 1 }$

$x = \dfrac { - 2 \pm \sqrt{ 4 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 6 } } { 2 \times 1 }$

$ $ Multiplying any number by 1 does not change the value $ $

$x = \dfrac { - 2 \pm \sqrt{ 4 - 4 \times 6 } } { 2 \times 1 }$

$x = \dfrac { - 2 \pm \sqrt{ 4 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } } } { 2 \times 1 }$

$ $ Multiply $ - 4 $ and $ 6$

$x = \dfrac { - 2 \pm \sqrt{ 4 \color{#FF6800}{ - } \color{#FF6800}{ 24 } } } { 2 \times 1 }$

$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 24 } } } { 2 \times 1 }$

$ $ Subtract $ 24 $ from $ 4$

$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 20 } } } { 2 \times 1 }$

$x = \dfrac { - 2 \pm \sqrt{ - 20 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$

$ $ Multiplying any number by 1 does not change the value $ $

$x = \dfrac { - 2 \pm \sqrt{ - 20 } } { \color{#FF6800}{ 2 } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ - 20 } } { 2 } }$

$ $ The square root of a negative number does not exist within the set of real numbers $ $

$ $ Do not have the solution $ $

$ $ Do not have the real root $ $

Find the number of solutions

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$

$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$

$\color{#FF6800}{ D } = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 }$

$D = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 6$

$ $ Calculate power $ $

$D = \color{#FF6800}{ 4 } - 4 \times 1 \times 6$

$D = 4 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 6$

$ $ Multiplying any number by 1 does not change the value $ $

$D = 4 - 4 \times 6$

$D = 4 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 }$

$ $ Multiply $ - 4 $ and $ 6$

$D = 4 \color{#FF6800}{ - } \color{#FF6800}{ 24 }$

$D = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 24 }$

$ $ Subtract $ 24 $ from $ 4$

$D = \color{#FF6800}{ - } \color{#FF6800}{ 20 }$

$\color{#FF6800}{ D } = \color{#FF6800}{ - } \color{#FF6800}{ 20 }$

$ $ Since $ D<0 $ , there is no real root of the following quadratic equation $ $

$ $ Do not have the real root $ $

$\alpha + \beta = - 2 , \alpha \beta = 6$

Find the sum and product of the two roots of the quadratic equation

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$

$ $ 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}$

$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 6 } { 1 } }$

$\alpha + \beta = - \dfrac { 2 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 6 } { 1 }$

$ $ If the denominator is 1, the denominator can be removed $ $

$\alpha + \beta = - \color{#FF6800}{ 2 } , \alpha \beta = \dfrac { 6 } { 1 }$

$\alpha + \beta = - 2 , \alpha \beta = \dfrac { 6 } { \color{#FF6800}{ 1 } }$

$ $ If the denominator is 1, the denominator can be removed $ $

$\alpha + \beta = - 2 , \alpha \beta = \color{#FF6800}{ 6 }$

$ $ 그래프 보기 $ $

Graph

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