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Formula
Solve the quadratic equation
Answer
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Solve the equation
Answer
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Number of solution
Answer
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Relationship between roots and coefficients
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Graph
$y = \left ( x - 1 \right ) \left ( x + 2 \right )$
$y = 0$
$x$-intercept
$\left ( 1 , 0 \right )$, $\left ( - 2 , 0 \right )$
$y$-intercept
$\left ( 0 , - 2 \right )$
Minimum
$\left ( - \dfrac { 1 } { 2 } , - \dfrac { 9 } { 4 } \right )$
Standard form
$y = \left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } - \dfrac { 9 } { 4 }$
$\left( x-1 \right) \left( x+2 \right) = 0$
$\begin{array} {l} x = 1 \\ x = - 2 \end{array}$
Find solution by method of factorization
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = 0$
$ $ Expand the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 1 \\ x = - 2 \end{array}$
Find out the solution of the equation of the multiplication form
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 2 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1 } - 4 \times 1 \times \left ( - 2 \right )$
$D = 1 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 2 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 1 - 4 \times \left ( - 2 \right )$
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$ $ Multiply $ - 4 $ and $ - 2$
$D = 1 + \color{#FF6800}{ 8 }$
$D = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 8 }$
$ $ Add $ 1 $ and $ 8$
$D = \color{#FF6800}{ 9 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 9 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 1 , \alpha \beta = - 2$
Find the sum and product of the two roots of the quadratic equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 2 } { 1 } }$
$\alpha + \beta = - \dfrac { 1 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 2 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 1 } , \alpha \beta = \dfrac { - 2 } { 1 }$
$\alpha + \beta = - 1 , \alpha \beta = \dfrac { - 2 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 1 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ 그래프 보기 $ $
Graph
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7th-9th grade
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7th-9th grade
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Calculus
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
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