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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
$y = x ^ { 2 } + 3 x$
$y = 0$
$x$Intercept
$\left ( 0 , 0 \right )$, $\left ( - 3 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( - \dfrac { 3 } { 2 } , - \dfrac { 9 } { 4 } \right )$
Standard form
$y = \left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } - \dfrac { 9 } { 4 }$
$x ^{ 2 } +3x = 0$
$\begin{array} {l} x = 0 \\ x = - 3 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = 0$
$ax^{2} + bx = x\left(ax+b\right)$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 0$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \end{array}$
$\begin{array} {l} x = 0 \\ x = - 3 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } = \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } = \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } = \dfrac { \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 3 ^ { 2 } } { 2 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 9 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 9 } { 4 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 9 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 9 } { 4 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \end{array}$
$\begin{array} {l} x = 0 \\ x = - 3 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = 0$
$ $ Bind the expressions with the common factor $ x$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 0$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ 3 ^ { 2 } - 4 \times 1 \times 0 } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ 9 } } { 2 \times 1 } }$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 9 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 3 \pm \color{#FF6800}{ 3 } } { 2 \times 1 }$
$x = \dfrac { - 3 \pm 3 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 3 \pm 3 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm 3 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 + 3 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 - 3 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } } { 2 } \\ x = \dfrac { - 3 - 3 } { 2 } \end{array}$
$ $ Remove the two numbers if the values are the same and the signs are different $ $
$\begin{array} {l} x = \dfrac { 0 } { 2 } \\ x = \dfrac { - 3 - 3 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 0 } { 2 } } \\ x = \dfrac { - 3 - 3 } { 2 } \end{array}$
$ $ If the numerator is 0, it is equal to 0 $ $
$\begin{array} {l} x = \color{#FF6800}{ 0 } \\ x = \dfrac { - 3 - 3 } { 2 } \end{array}$
$\begin{array} {l} x = 0 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 2 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 0 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } } { 2 } \end{array}$
$\begin{array} {l} x = 0 \\ x = \color{#FF6800}{ \dfrac { - 6 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 0 \\ x = \color{#FF6800}{ \dfrac { - 3 } { 1 } } \end{array}$
$\begin{array} {l} x = 0 \\ x = \dfrac { - 3 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 0 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$D = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 0$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 9 } - 4 \times 1 \times 0$
$D = 9 - 4 \times 1 \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$ $ If you multiply a number by 0, it becomes 0 $ $
$D = 9 + \color{#FF6800}{ 0 }$
$D = 9 \color{#FF6800}{ + } \color{#FF6800}{ 0 }$
$ $ 0 does not change when you add or subtract $ $
$D = 9$
$\color{#FF6800}{ D } = \color{#FF6800}{ 9 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 3 , \alpha \beta = 0$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 0 } { 1 } }$
$\alpha + \beta = - \dfrac { 3 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 0 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 3 } , \alpha \beta = \dfrac { 0 } { 1 }$
$\alpha + \beta = - 3 , \alpha \beta = \dfrac { 0 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 3 , \alpha \beta = \color{#FF6800}{ 0 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-Solve . 
$2-10P$ $+21=0$ 
$D2$ 
$x^{2+x-}$ $12=0$ 
$0=0$ 
$2x^{2}+3x=20$ 
$2a^{2}+5a+3=0$ 
$3a^{2}+7a+6=0$ 
$⑤$ 
$\left(6\right)$ $x^{2}+6x+5=0$ 
$x^{2}+3x-28=°$ 

$-5x-14=0$ 
$8$ $x^{2}$ $-5x$ 
$6x^{2}-11x+4=0$ 
$3x^{2}+2x-8=0$
7th-9th grade
Other
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es and shade only the letter that corresponds 
to your answer. 
Which of the following quadratic equation is written in standard $o$ $7$ 
a. $3x-2x^{2}=7$ b. $2x^{2}-7=-3x$ c. $-2x^{2}-7+3x=0$ d. 
in $sandard$ form of a quadratic equation. 
Write $0\left(x+3\left(x+4\right)=0$ $x^{2}+3x+4$ b. $x^{2}+6x+12$ $C$ C. $x^{2}+7x+12$ 
a. 
Which of the following quadratic equation $\square $ in a $60$ $ax^{2}+c=07$ 
$a$ $2x^{2}+3x=0$ b. $2x^{2}=3x$ $C$ $2x^{2}+3=5$
10th-13th grade
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