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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
$y = x \left ( x + 15 \right )$
$y = 5 \left ( x - 5 \right )$
$x$Intercept
$\left ( - 15 , 0 \right )$, $\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( - \dfrac { 15 } { 2 } , - \dfrac { 225 } { 4 } \right )$
Standard form
$y = \left ( x + \dfrac { 15 } { 2 } \right ) ^ { 2 } - \dfrac { 225 } { 4 }$
$x$Intercept
$\left ( 5 , 0 \right )$
$y$Intercept
$\left ( 0 , - 25 \right )$
$x \left( x+15 \right) = 5 \left( x-5 \right)$
$x = - 5$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 5 \left ( x - 5 \right )$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ x } = 5 \left ( x - 5 \right )$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$ $ Organize the expression $ $
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 25 }$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } - 25$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + 15 x \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = 0$
$x ^ { 2 } + \color{#FF6800}{ 15 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } + 25 = 0$
$ $ Calculate between similar terms $ $
$x ^ { 2 } + \color{#FF6800}{ 10 } \color{#FF6800}{ x } + 25 = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = 0$
$ $ Express as the perfect square formula $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$x = - 5$
Calculate using the quadratic formula
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 5 \left ( x - 5 \right )$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ x } = 5 \left ( x - 5 \right )$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$ $ Organize the expression $ $
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 25 }$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } - 25$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + 15 x \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = \color{#FF6800}{ 0 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 10 \pm \sqrt{ 0 } } { 2 \times 1 } }$
$x = \dfrac { - 10 \pm \sqrt{ \color{#FF6800}{ 0 } } } { 2 \times 1 }$
$n square root $ of 0 is 0 $ $
$x = \dfrac { - 10 \pm \color{#FF6800}{ 0 } } { 2 \times 1 }$
$x = \dfrac { - 10 \pm 0 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 10 \pm 0 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 10 \pm 0 } { 2 } }$
$ $ The value will not be changed even if adding or subtracting 0 $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 10 } { 2 } }$
$x = \color{#FF6800}{ \dfrac { - 10 } { 2 } }$
$ $ Do the reduction of the fraction format $ $
$x = \color{#FF6800}{ \dfrac { - 5 } { 1 } }$
$x = \dfrac { - 5 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$x = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$ $ 1 real root (multiple root) $ $
Find the number of solutions
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 5 \left ( x - 5 \right )$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ x } = 5 \left ( x - 5 \right )$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$ $ Organize the expression $ $
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 25 }$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } - 25$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + 15 x \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = 0$
$x ^ { 2 } + \color{#FF6800}{ 15 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } + 25 = 0$
$ $ Calculate between similar terms $ $
$x ^ { 2 } + \color{#FF6800}{ 10 } \color{#FF6800}{ x } + 25 = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 10 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 25 }$
$D = \color{#FF6800}{ 10 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 25$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 100 } - 4 \times 1 \times 25$
$D = 100 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 25$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 100 - 4 \times 25$
$D = 100 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 25 }$
$ $ Multiply $ - 4 $ and $ 25$
$D = 100 \color{#FF6800}{ - } \color{#FF6800}{ 100 }$
$D = \color{#FF6800}{ 100 } \color{#FF6800}{ - } \color{#FF6800}{ 100 }$
$ $ Remove the two numbers if the values are the same and the signs are different $ $
$D = 0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 }$
$ $ Since $ D=0 $ , the number of real root of the following quadratic equation is 1 (multiple root) $ $
$ $ 1 real root (multiple root) $ $
$\alpha + \beta = - 10 , \alpha \beta = 25$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 5 \left ( x - 5 \right )$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ x } = 5 \left ( x - 5 \right )$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$ $ Organize the expression $ $
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 25 }$
$x ^ { 2 } + 15 x = \color{#FF6800}{ 5 } \color{#FF6800}{ x } - 25$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + 15 x \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = 0$
$x ^ { 2 } + \color{#FF6800}{ 15 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } + 25 = 0$
$ $ Calculate between similar terms $ $
$x ^ { 2 } + \color{#FF6800}{ 10 } \color{#FF6800}{ x } + 25 = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 25 } { 1 } }$
$\alpha + \beta = - \dfrac { 10 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 25 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 10 } , \alpha \beta = \dfrac { 25 } { 1 }$
$\alpha + \beta = - 10 , \alpha \beta = \dfrac { 25 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 10 , \alpha \beta = \color{#FF6800}{ 25 }$
Solution search results
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