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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
$y = x ^ { 2 } - 4 x + 3$
$y = 0$
$x$Intercept
$\left ( 3 , 0 \right )$, $\left ( 1 , 0 \right )$
$y$Intercept
$\left ( 0 , 3 \right )$
Minimum
$\left ( 2 , - 1 \right )$
Standard form
$y = \left ( x - 2 \right ) ^ { 2 } - 1$
$x ^{ 2 } -4x+3 = 0$
$\begin{array} {l} x = 3 \\ x = 1 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = 3 \\ x = 1 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - 2 \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - 2 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x - 2 \right ) ^ { 2 } = - 3 + \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } }$
$ $ Calculate power $ $
$\left ( x - 2 \right ) ^ { 2 } = - 3 + \color{#FF6800}{ 4 }$
$\left ( x - 2 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
$ $ Add $ - 3 $ and $ 4$
$\left ( x - 2 \right ) ^ { 2 } = \color{#FF6800}{ 1 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ x = 2 - 1 \end{array}$
$ $ Add $ 2 $ and $ 1$
$\begin{array} {l} x = \color{#FF6800}{ 3 } \\ x = 2 - 1 \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$ $ Subtract $ 1 $ from $ 2$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = 3 \\ x = 1 \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 4 \right ) \pm \sqrt{ \left ( - 4 \right ) ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 4 \pm \sqrt{ \left ( - 4 \right ) ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 }$
$x = \dfrac { 4 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 3 } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 4 \pm \sqrt{ 4 ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 \pm \sqrt{ 4 ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 \pm \sqrt{ 4 } } { 2 \times 1 } }$
$x = \dfrac { 4 \pm \sqrt{ \color{#FF6800}{ 4 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 4 \pm \color{#FF6800}{ 2 } } { 2 \times 1 }$
$x = \dfrac { 4 \pm 2 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 4 \pm 2 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 \pm 2 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 + 2 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 - 2 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } } { 2 } \\ x = \dfrac { 4 - 2 } { 2 } \end{array}$
$ $ Add $ 4 $ and $ 2$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 6 } } { 2 } \\ x = \dfrac { 4 - 2 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 } { 2 } } \\ x = \dfrac { 4 - 2 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \\ x = \dfrac { 4 - 2 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \\ x = \dfrac { 4 - 2 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 3 } \\ x = \dfrac { 4 - 2 } { 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 2 } \end{array}$
$ $ Subtract $ 2 $ from $ 4$
$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ 2 } } { 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { 2 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { 1 } { 1 } } \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { 1 } { 1 } } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ 1 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 3$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 4 ^ { 2 } - 4 \times 1 \times 3$
$D = \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 3$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 16 } - 4 \times 1 \times 3$
$D = 16 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 3$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 16 - 4 \times 3$
$D = 16 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 }$
$ $ Multiply $ - 4 $ and $ 3$
$D = 16 \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$D = \color{#FF6800}{ 16 } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$ $ Subtract $ 12 $ from $ 16$
$D = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 4 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 4 , \alpha \beta = 3$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 4 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 3 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 4 } { 1 } } , \alpha \beta = \dfrac { 3 } { 1 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 4 } { 1 } } , \alpha \beta = \dfrac { 3 } { 1 }$
$\alpha + \beta = \dfrac { 4 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 3 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = \color{#FF6800}{ 4 } , \alpha \beta = \dfrac { 3 } { 1 }$
$\alpha + \beta = 4 , \alpha \beta = \dfrac { 3 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 4 , \alpha \beta = \color{#FF6800}{ 3 }$
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