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Formula

Solve the quadratic equation

Answer

Number of solution

Answer

Relationship between roots and coefficients

Answer

Graph

$y = 5 x ^ { 2 } - 12$

$y = 0$

$x$Intercept

$\left ( - \dfrac { 2 \sqrt{ 15 } } { 5 } , 0 \right )$, $\left ( \dfrac { 2 \sqrt{ 15 } } { 5 } , 0 \right )$

$y$Intercept

$\left ( 0 , - 12 \right )$

Minimum

$\left ( 0 , - 12 \right )$

Standard form

$y = 5 x ^ { 2 } - 12$

$5x ^{ 2 } -12 = 0$

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = - \dfrac { 2 \sqrt{ 15 } } { 5 } \end{array}$

Solve quadratic equations using the square root

$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$

$ $ Divide both sides by the coefficient of the leading highest term $ $

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } } = \color{#FF6800}{ 0 }$

$x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } } = 0$

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

$x ^ { 2 } = \color{#FF6800}{ \dfrac { 12 } { 5 } }$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 12 } { 5 } }$

$ $ Solve quadratic equations using the square root $ $

$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 12 } { 5 } } }$

$ $ Solve a solution to $ x$

$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \end{array}$

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = - \dfrac { 2 \sqrt{ 15 } } { 5 } \end{array}$

Calculate using the quadratic formula

$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$

$ $ 0 has no sign $ $

$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$

$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$

$ $ The power of 0 is 0 $ $

$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 0 - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 } }$

$ $ Organize the expression $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 240 } } { 2 \times 5 } }$

$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 240 } } } { 2 \times 5 }$

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

$x = \dfrac { 0 \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 15 } } } { 2 \times 5 }$

$x = \dfrac { 0 \pm 4 \sqrt{ 15 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } }$

$ $ Multiply $ 2 $ and $ 5$

$x = \dfrac { 0 \pm 4 \sqrt{ 15 } } { \color{#FF6800}{ 10 } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm 4 \sqrt{ 15 } } { 10 } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 4 \sqrt{ 15 } } { 10 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } } \end{array}$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 0 } + 4 \sqrt{ 15 } } { 10 } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$

$ $ 0 does not change when you add or subtract $ $

$\begin{array} {l} x = \dfrac { 4 \sqrt{ 15 } } { 10 } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 4 \sqrt{ 15 } } { 10 } } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \dfrac { \color{#FF6800}{ 0 } - 4 \sqrt{ 15 } } { 10 } \end{array}$

$ $ 0 does not change when you add or subtract $ $

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \dfrac { - 4 \sqrt{ 15 } } { 10 } \end{array}$

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ \dfrac { - 4 \sqrt{ 15 } } { 10 } } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 15 } } { 5 } } \end{array}$

$ $ Move the minus sign to the front of the fraction $ $

$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \end{array}$

$ $ 2 real roots $ $

Find the number of solutions

$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$

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

$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$

$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times \left ( - 12 \right )$

$ $ The power of 0 is 0 $ $

$D = \color{#FF6800}{ 0 } - 4 \times 5 \times \left ( - 12 \right )$

$ $ 0 does not change when you add or subtract $ $

$D = - 4 \times 5 \times \left ( - 12 \right )$

$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$

$ $ Multiply the numbers $ $

$D = \color{#FF6800}{ 240 }$

$\color{#FF6800}{ D } = \color{#FF6800}{ 240 }$

$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $

$ $ 2 real roots $ $

$\alpha + \beta = 0 , \alpha \beta = - \dfrac { 12 } { 5 }$

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

$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$

$ $ 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}$

$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 5 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 12 } { 5 } }$

$\alpha + \beta = - \color{#FF6800}{ \dfrac { 0 } { 5 } } , \alpha \beta = \dfrac { - 12 } { 5 }$

$ $ If the numerator is 0, it is equal to 0 $ $

$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 5 }$

$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 5 }$

$ $ 0 has no sign $ $

$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 5 }$

$\alpha + \beta = 0 , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 12 } } { 5 }$

$ $ Move the minus sign to the front of the fraction $ $

$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } }$

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