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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
$y = 3 x ^ { 2 } + 4 x$
$y = 0$
$x$Intercept
$\left ( - \dfrac { 4 } { 3 } , 0 \right )$, $\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( - \dfrac { 2 } { 3 } , - \dfrac { 4 } { 3 } \right )$
Standard form
$y = 3 \left ( x + \dfrac { 2 } { 3 } \right ) ^ { 2 } - \dfrac { 4 } { 3 }$
$3x ^{ 2 } +4x = 0$
$\begin{array} {l} x = 0 \\ x = - \dfrac { 4 } { 3 } \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } = 0$
$ax^{2} + bx = x\left(ax+b\right)$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) = 0$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \end{array}$
$\begin{array} {l} x = 0 \\ x = - \dfrac { 4 } { 3 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 2 } { 3 } \right ) ^ { 2 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 2 } { 3 } \right ) ^ { 2 } = \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 2 } { 3 } \right ) ^ { 2 } = \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 2 } { 3 } \right ) ^ { 2 } = \dfrac { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 2 ^ { 2 } } { 3 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 4 } { 9 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 4 } { 9 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 4 } { 9 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 4 } { 9 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \end{array}$
$\begin{array} {l} x = 0 \\ x = - \dfrac { 4 } { 3 } \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } = 0$
$ $ Bind the expressions with the common factor $ x$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) = 0$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 4 \pm \sqrt{ 4 ^ { 2 } - 4 \times 3 \times 0 } } { 2 \times 3 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 4 \pm \sqrt{ 16 } } { 2 \times 3 } }$
$x = \dfrac { - 4 \pm \sqrt{ \color{#FF6800}{ 16 } } } { 2 \times 3 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 4 \pm \color{#FF6800}{ 4 } } { 2 \times 3 }$
$x = \dfrac { - 4 \pm 4 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { - 4 \pm 4 } { \color{#FF6800}{ 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 4 \pm 4 } { 6 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 4 + 4 } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 4 - 4 } { 6 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } } { 6 } \\ x = \dfrac { - 4 - 4 } { 6 } \end{array}$
$ $ Remove the two numbers if the values are the same and the signs are different $ $
$\begin{array} {l} x = \dfrac { 0 } { 6 } \\ x = \dfrac { - 4 - 4 } { 6 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 0 } { 6 } } \\ x = \dfrac { - 4 - 4 } { 6 } \end{array}$
$ $ If the numerator is 0, it is equal to 0 $ $
$\begin{array} {l} x = \color{#FF6800}{ 0 } \\ x = \dfrac { - 4 - 4 } { 6 } \end{array}$
$\begin{array} {l} x = 0 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } } { 6 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 0 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } } { 6 } \end{array}$
$\begin{array} {l} x = 0 \\ x = \color{#FF6800}{ \dfrac { - 8 } { 6 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 0 \\ x = \color{#FF6800}{ \dfrac { - 4 } { 3 } } \end{array}$
$\begin{array} {l} x = 0 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 4 } } { 3 } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = 0 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$D = \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 0$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 16 } - 4 \times 3 \times 0$
$D = 16 - 4 \times 3 \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$ $ If you multiply a number by 0, it becomes 0 $ $
$D = 16 + \color{#FF6800}{ 0 }$
$D = 16 \color{#FF6800}{ + } \color{#FF6800}{ 0 }$
$ $ 0 does not change when you add or subtract $ $
$D = 16$
$\color{#FF6800}{ D } = \color{#FF6800}{ 16 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - \dfrac { 4 } { 3 } , \alpha \beta = 0$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 } { 3 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 0 } { 3 } }$
$\alpha + \beta = - \dfrac { 4 } { 3 } , \alpha \beta = \color{#FF6800}{ \dfrac { 0 } { 3 } }$
$ $ If the numerator is 0, it is equal to 0 $ $
$\alpha + \beta = - \dfrac { 4 } { 3 } , \alpha \beta = \color{#FF6800}{ 0 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$14$ Which quadratic equation has $1$ as the sum of its roots? 
$A$ $x^{2}+4x-4=0$ $3$ $x^{2}-4x-4=0$ C $x^{2}+4x+4=0$ $\square $ $x^{2}+4x=0$ 
$15$ Whuch equation is transformable to quadratic equation? 
$A$ $3\left(x+2\right)=4$ $B$ $n\left(n+1\right)=4$ C. $3=t^{2}\left(2-1\right)$ $\square $ D $\left(z^{2}+1\right)z^{2}+8\right)=0$
7th-9th grade
Other
search-thumbnail-$8.$ $3x^{2}-4x=0$ 
$9.$ $9x^{2}-72=0$ 
$10.$ $2x^{2}+4x=3$
7th-9th grade
Other
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