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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
$y = \dfrac { 1 } { 5 } x ^ { 2 } - \dfrac { 2 } { 5 } x - \dfrac { 1 } { 3 }$
$y = 0$
$x$-intercept
$\left ( 1 - \dfrac { 2 \sqrt{ 6 } } { 3 } , 0 \right )$, $\left ( 1 + \dfrac { 2 \sqrt{ 6 } } { 3 } , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 1 } { 3 } \right )$
Minimum
$\left ( 1 , - \dfrac { 8 } { 15 } \right )$
Standard form
$y = \dfrac { 1 } { 5 } \left ( x - 1 \right ) ^ { 2 } - \dfrac { 8 } { 15 }$
$\dfrac{ 1 }{ 5 } x ^{ 2 } - \dfrac{ 2 }{ 5 } x- \dfrac{ 1 }{ 3 } = 0$
$\begin{array} {l} x = \dfrac { 3 + 2 \sqrt{ 6 } } { 3 } \\ x = \dfrac { 3 - 2 \sqrt{ 6 } } { 3 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = 0$
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 8 } { 3 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 8 } { 3 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 8 } { 3 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 8 } { 3 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 3 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 3 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 + 2 \sqrt{ 6 } } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 - 2 \sqrt{ 6 } } { 3 } } \end{array}$
$\begin{array} {l} x = \dfrac { 3 + 2 \sqrt{ 6 } } { 3 } \\ x = \dfrac { 3 - 2 \sqrt{ 6 } } { 3 } \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = 0$
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 6 \right ) \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 3 \times \left ( - 5 \right ) } } { 2 \times 3 }$
$ $ Simplify Minus $ $
$x = \dfrac { 6 \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 3 \times \left ( - 5 \right ) } } { 2 \times 3 }$
$x = \dfrac { 6 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times \left ( - 5 \right ) } } { 2 \times 3 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 3 \times \left ( - 5 \right ) } } { 2 \times 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 3 \times \left ( - 5 \right ) } } { 2 \times 3 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm \sqrt{ 96 } } { 2 \times 3 } }$
$x = \dfrac { 6 \pm \sqrt{ \color{#FF6800}{ 96 } } } { 2 \times 3 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 6 \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 6 } } } { 2 \times 3 }$
$x = \dfrac { 6 \pm 4 \sqrt{ 6 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { 6 \pm 4 \sqrt{ 6 } } { \color{#FF6800}{ 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm 4 \sqrt{ 6 } } { 6 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 + 4 \sqrt{ 6 } } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 - 4 \sqrt{ 6 } } { 6 } } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 + 4 \sqrt{ 6 } } { 6 } } \\ x = \dfrac { 6 - 4 \sqrt{ 6 } } { 6 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 + 2 \sqrt{ 6 } } { 3 } } \\ x = \dfrac { 6 - 4 \sqrt{ 6 } } { 6 } \end{array}$
$\begin{array} {l} x = \dfrac { 3 + 2 \sqrt{ 6 } } { 3 } \\ x = \color{#FF6800}{ \dfrac { 6 - 4 \sqrt{ 6 } } { 6 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \dfrac { 3 + 2 \sqrt{ 6 } } { 3 } \\ x = \color{#FF6800}{ \dfrac { 3 - 2 \sqrt{ 6 } } { 3 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = 0$
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times \left ( - 5 \right )$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 6 ^ { 2 } - 4 \times 3 \times \left ( - 5 \right )$
$D = \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times \left ( - 5 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 36 } - 4 \times 3 \times \left ( - 5 \right )$
$D = 36 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$ $ Multiply the numbers $ $
$D = 36 + \color{#FF6800}{ 60 }$
$D = \color{#FF6800}{ 36 } \color{#FF6800}{ + } \color{#FF6800}{ 60 }$
$ $ Add $ 36 $ and $ 60$
$D = \color{#FF6800}{ 96 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 96 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 2 , \alpha \beta = - \dfrac { 5 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = 0$
$\color{#FF6800}{ \dfrac { 3 x ^ { 2 } - 6 x - 5 } { 15 } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 6 } { 3 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 5 } { 3 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 6 } { 3 } } , \alpha \beta = \dfrac { - 5 } { 3 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 6 } { 3 } } , \alpha \beta = \dfrac { - 5 } { 3 }$
$\alpha + \beta = \color{#FF6800}{ \dfrac { 6 } { 3 } } , \alpha \beta = \dfrac { - 5 } { 3 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = \color{#FF6800}{ 2 } , \alpha \beta = \dfrac { - 5 } { 3 }$
$\alpha + \beta = 2 , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { 3 }$
$ $ Move the minus sign to the front of the fraction $ $
$\alpha + \beta = 2 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$24$ $\int \dfrac {1} {\sqrt{x} }\left(1+x\right)\left(1+x+x^{2}\right)dx=$ 
(A) $\right)$ $2\sqrt{x} \left(\dfrac {1} {7}x^{3}+\dfrac {1} {5}x^{2}+\dfrac {1} {3}x+1\right)+c$ 
$\left(B\right)$ $2\sqrt{x} \left(\dfrac {1} {7}x^{3}+\dfrac {2} {5}x^{2}+\dfrac {2} {3}x+1\right)+c$ 
(C) $2\sqrt{x} \left(\dfrac {2} {7}x^{3}+\dfrac {2} {5}x^{2}+\dfrac {2} {3}x+1\right)+c$ 
(D) $2\sqrt{x} \left(\dfrac {2} {7}x^{3}+\dfrac {4} {5}x^{2}+\dfrac {4} {3}x+1\right)+c$ 
(E) $2\sqrt{x} \left(\dfrac {2} {7}x^{3}+\dfrac {4} {5}x^{2}+\dfrac {4} {3}x+2\right)+c$
10th-13th grade
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