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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
$y = 0.3 x ^ { 2 } - 0.6 x$
$y = 1$
$x$Intercept
$\left ( 2 , 0 \right )$, $\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( 1 , - \dfrac { 3 } { 10 } \right )$
Standard form
$y = \dfrac { 3 } { 10 } \left ( x - 1 \right ) ^ { 2 } - \dfrac { 3 } { 10 }$
$\begin{array} {l} x = \dfrac { 3 + \sqrt{ 39 } } { 3 } \\ x = \dfrac { 3 - \sqrt{ 39 } } { 3 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 0.3 } x ^ { 2 } - 0.6 x = 1$
$ $ Convert decimals to fractions $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } x ^ { 2 } - 0.6 x = 1$
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 0.6 } x = 1$
$ $ Convert decimals to fractions $ $
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } x = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } \color{#FF6800}{ x } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 }$
$3 x ^ { 2 } - 6 x = \color{#FF6800}{ 10 }$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - 6 x \color{#FF6800}{ - } \color{#FF6800}{ 10 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 10 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 10 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 10 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 10 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 3 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 3 } } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 3 } } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 3 } } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \end{array}$
$\begin{array} {l} x = \dfrac { 3 + \sqrt{ 39 } } { 3 } \\ x = \dfrac { 3 - \sqrt{ 39 } } { 3 } \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ 0.3 } x ^ { 2 } - 0.6 x = 1$
$ $ Convert decimals to fractions $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } x ^ { 2 } - 0.6 x = 1$
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 0.6 } x = 1$
$ $ Convert decimals to fractions $ $
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } x = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } \color{#FF6800}{ x } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 }$
$3 x ^ { 2 } - 6 x = \color{#FF6800}{ 10 }$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - 6 x \color{#FF6800}{ - } \color{#FF6800}{ 10 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } } }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 6 \right ) \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 3 \times \left ( - 10 \right ) } } { 2 \times 3 }$
$ $ Simplify Minus $ $
$x = \dfrac { 6 \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 3 \times \left ( - 10 \right ) } } { 2 \times 3 }$
$x = \dfrac { 6 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times \left ( - 10 \right ) } } { 2 \times 3 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 3 \times \left ( - 10 \right ) } } { 2 \times 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \pm \sqrt{ \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \pm \sqrt{ \color{#FF6800}{ 156 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } } }$
$x = \dfrac { 6 \pm \sqrt{ \color{#FF6800}{ 156 } } } { 2 \times 3 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 6 \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 39 } } } { 2 \times 3 }$
$x = \dfrac { 6 \pm 2 \sqrt{ 39 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { 6 \pm 2 \sqrt{ 39 } } { \color{#FF6800}{ 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 6 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 6 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 6 } } } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 6 } } } \\ x = \dfrac { 6 - 2 \sqrt{ 39 } } { 6 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \\ x = \dfrac { 6 - 2 \sqrt{ 39 } } { 6 } \end{array}$
$\begin{array} {l} x = \dfrac { 3 + \sqrt{ 39 } } { 3 } \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 6 } } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \dfrac { 3 + \sqrt{ 39 } } { 3 } \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 39 } } } { \color{#FF6800}{ 3 } } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 0.3 } x ^ { 2 } - 0.6 x = 1$
$ $ Convert decimals to fractions $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } x ^ { 2 } - 0.6 x = 1$
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 0.6 } x = 1$
$ $ Convert decimals to fractions $ $
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } x = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } \color{#FF6800}{ x } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 }$
$3 x ^ { 2 } - 6 x = \color{#FF6800}{ 10 }$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - 6 x \color{#FF6800}{ - } \color{#FF6800}{ 10 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times \left ( - 10 \right )$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 6 ^ { 2 } - 4 \times 3 \times \left ( - 10 \right )$
$D = \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times \left ( - 10 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 36 } - 4 \times 3 \times \left ( - 10 \right )$
$D = 36 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right )$
$ $ Multiply the numbers $ $
$D = 36 + \color{#FF6800}{ 120 }$
$D = \color{#FF6800}{ 36 } \color{#FF6800}{ + } \color{#FF6800}{ 120 }$
$ $ Add $ 36 $ and $ 120$
$D = \color{#FF6800}{ 156 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 156 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 2 , \alpha \beta = - \dfrac { 10 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 0.3 } x ^ { 2 } - 0.6 x = 1$
$ $ Convert decimals to fractions $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } x ^ { 2 } - 0.6 x = 1$
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 0.6 } x = 1$
$ $ Convert decimals to fractions $ $
$\dfrac { 3 } { 10 } x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } x = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 10 } } } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 5 } } } \color{#FF6800}{ x } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = 1$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } } { \color{#FF6800}{ 10 } } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 }$
$3 x ^ { 2 } - 6 x = \color{#FF6800}{ 10 }$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - 6 x \color{#FF6800}{ - } \color{#FF6800}{ 10 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } } { \color{#FF6800}{ 3 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { \color{#FF6800}{ 3 } } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } } { \color{#FF6800}{ 3 } } } , \alpha \beta = \dfrac { - 10 } { 3 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 3 } } } , \alpha \beta = \dfrac { - 10 } { 3 }$
$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 3 } } } , \alpha \beta = \dfrac { - 10 } { 3 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = \color{#FF6800}{ 2 } , \alpha \beta = \dfrac { - 10 } { 3 }$
$\alpha + \beta = 2 , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 3 }$
$ $ Move the minus sign to the front of the fraction $ $
$\alpha + \beta = 2 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 10 } } { \color{#FF6800}{ 3 } } }$
Solution search results
search-thumbnail-11)\dfrac{m}{4}=-1
$11\right)\dfrac {m} {4}=-1$ $13\right)$ $20=15+y$ $15\right)$ $-0.6x=11.22$ $17\right)\dfrac {n} {8.2}=-3.7$
7th-9th grade
Algebra
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search-thumbnail-∠ F From the sum ot x^{A}(2)-8 with 3x-12 subtract the sum ofx-9
$∠$ $F$ From the sum $ot$ $x^{A}\left(2\right)-8$ with $3x-12$ subtract the sum $ofx-9$ with $3x-x^{A}\left(2\right)$ Multinlication takes place $\left(2\times A$ $\left(2\right)+5x+21\right)\left(x-$ $5\right)=$ #Please provide solution by using math formula.
10th-13th grade
Geometry
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search-thumbnail-e 6x=12
$e$ $6x=12$
1st-6th grade
Other
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search-thumbnail-5. 6x=12
$5.$ $6x=12$ $1$
10th-13th grade
Calculus
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search-thumbnail-Using the \emph{removal of first derivative}
Using the \emph{removal of first derivative} method, the differential equation \( \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as \). For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$
Calculus
Search count: 3,465
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search-thumbnail- 31 x=3 1 y 2
$x^{2}+y^{2}-6x+2y+5=0$ $31$ $x=3$ $1$ $y$ $2$ $y^{2}-x^{2}=k^{2}$ $2^{1}$ $γ$
10th-13th grade
Other
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