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Formula
Solve the quadratic equation
Answer
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Number of solution
Answer
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Relationship between roots and coefficients
Answer
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Graph
$y = 5 x ^ { 2 } - 10 x$
$y = 0$
$x$Intercept
$\left ( 2 , 0 \right )$, $\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( 1 , - 5 \right )$
Standard form
$y = 5 \left ( x - 1 \right ) ^ { 2 } - 5$
$5x ^{ 2 } -10x = 0$
$\begin{array} {l} x = 0 \\ x = 2 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ x } = 0$
$ax^{2} + bx = x\left(ax+b\right)$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 2 \\ x = 0 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - 1 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - 1 \right ) ^ { 2 } = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x - 1 \right ) ^ { 2 } = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
$ $ Calculate power $ $
$\left ( x - 1 \right ) ^ { 2 } = \color{#FF6800}{ 1 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ x = 1 - 1 \end{array}$
$ $ Add $ 1 $ and $ 1$
$\begin{array} {l} x = \color{#FF6800}{ 2 } \\ x = 1 - 1 \end{array}$
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$ $ Remove the two numbers if the values are the same and the signs are different $ $
$\begin{array} {l} x = 2 \\ x = 0 \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 0$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 10 ^ { 2 } - 4 \times 5 \times 0$
$D = \color{#FF6800}{ 10 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 0$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 100 } - 4 \times 5 \times 0$
$D = 100 - 4 \times 5 \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$ $ If you multiply a number by 0, it becomes 0 $ $
$D = 100 + \color{#FF6800}{ 0 }$
$D = 100 \color{#FF6800}{ + } \color{#FF6800}{ 0 }$
$ $ 0 does not change when you add or subtract $ $
$D = 100$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 2 , \alpha \beta = 0$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 10 } { 5 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 0 } { 5 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 10 } { 5 } } , \alpha \beta = \dfrac { 0 } { 5 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 10 } { 5 } } , \alpha \beta = \dfrac { 0 } { 5 }$
$\alpha + \beta = \color{#FF6800}{ \dfrac { 10 } { 5 } } , \alpha \beta = \dfrac { 0 } { 5 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = \color{#FF6800}{ 2 } , \alpha \beta = \dfrac { 0 } { 5 }$
$\alpha + \beta = 2 , \alpha \beta = \color{#FF6800}{ \dfrac { 0 } { 5 } }$
$ $ If the numerator is 0, it is equal to 0 $ $
$\alpha + \beta = 2 , \alpha \beta = \color{#FF6800}{ 0 }$
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