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Solve the quadratic equation
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Number of solution
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Relationship between roots and coefficients
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Graph
$y = 40 x - 5 x ^ { 2 }$
$y = 75$
$x$Intercept
$\left ( 8 , 0 \right )$, $\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Maximum
$\left ( 4 , 80 \right )$
Standard form
$y = - 5 \left ( x - 4 \right ) ^ { 2 } + 80$
$\begin{array} {l} x = 5 \\ x = 3 \end{array}$
Find solution by method of factorization
$40 x - 5 x ^ { 2 } = \color{#FF6800}{ 75 }$
$ $ Move the expression to the left side and change the symbol $ $
$40 x - 5 x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$ $ Expand the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 3 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
$ $ Move the expression to the left side and change the symbol $ $
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 3 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
$ $ Move the expression to the left side and change the symbol $ $
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 75 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } } }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 40 \right ) \pm \sqrt{ \left ( - 40 \right ) ^ { 2 } - 4 \times 5 \times 75 } } { 2 \times 5 }$
$ $ Simplify Minus $ $
$x = \dfrac { 40 \pm \sqrt{ \left ( - 40 \right ) ^ { 2 } - 4 \times 5 \times 75 } } { 2 \times 5 }$
$x = \dfrac { 40 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 75 } } { 2 \times 5 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 40 \pm \sqrt{ 40 ^ { 2 } - 4 \times 5 \times 75 } } { 2 \times 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } \pm \sqrt{ \color{#FF6800}{ 40 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 75 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } \pm \sqrt{ \color{#FF6800}{ 100 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } } }$
$x = \dfrac { 40 \pm \sqrt{ \color{#FF6800}{ 100 } } } { 2 \times 5 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 40 \pm \color{#FF6800}{ 10 } } { 2 \times 5 }$
$x = \dfrac { 40 \pm 10 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } }$
$ $ Multiply $ 2 $ and $ 5$
$x = \dfrac { 40 \pm 10 } { \color{#FF6800}{ 10 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } \pm \color{#FF6800}{ 10 } } { \color{#FF6800}{ 10 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } } { \color{#FF6800}{ 10 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { \color{#FF6800}{ 10 } } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 40 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } } { 10 } \\ x = \dfrac { 40 - 10 } { 10 } \end{array}$
$ $ Add $ 40 $ and $ 10$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 50 } } { 10 } \\ x = \dfrac { 40 - 10 } { 10 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 50 } } { \color{#FF6800}{ 10 } } } \\ x = \dfrac { 40 - 10 } { 10 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { 40 - 10 } { 10 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { 40 - 10 } { 10 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 5 } \\ x = \dfrac { 40 - 10 } { 10 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 40 } \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 10 } \end{array}$
$ $ Subtract $ 10 $ from $ 40$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 30 } } { 10 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } } { \color{#FF6800}{ 10 } } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 1 } } } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 1 } } } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ 3 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
$ $ Move the expression to the left side and change the symbol $ $
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 75 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 75$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 40 ^ { 2 } - 4 \times 5 \times 75$
$D = \color{#FF6800}{ 40 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 75$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1600 } - 4 \times 5 \times 75$
$D = 1600 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 75 }$
$ $ Multiply the numbers $ $
$D = 1600 \color{#FF6800}{ - } \color{#FF6800}{ 1500 }$
$D = \color{#FF6800}{ 1600 } \color{#FF6800}{ - } \color{#FF6800}{ 1500 }$
$ $ Subtract $ 1500 $ from $ 1600$
$D = \color{#FF6800}{ 100 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 8 , \alpha \beta = 15$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
$ $ Move the expression to the left side and change the symbol $ $
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 40 } } { \color{#FF6800}{ 5 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 75 } } { \color{#FF6800}{ 5 } } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 40 } } { \color{#FF6800}{ 5 } } } , \alpha \beta = \dfrac { 75 } { 5 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } } { \color{#FF6800}{ 5 } } } , \alpha \beta = \dfrac { 75 } { 5 }$
$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 40 } } { \color{#FF6800}{ 5 } } } , \alpha \beta = \dfrac { 75 } { 5 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = \color{#FF6800}{ 8 } , \alpha \beta = \dfrac { 75 } { 5 }$
$\alpha + \beta = 8 , \alpha \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 75 } } { \color{#FF6800}{ 5 } } }$
$ $ Reduce the fraction $ $
$\alpha + \beta = 8 , \alpha \beta = \color{#FF6800}{ 15 }$
Solution search results
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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
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search-thumbnail-4x^{8} -10x^{6}-5x^{4}\div2x^{3}
$0$ $4x^{8}$ $-10x$ $-10x^{6}-5x^{4}\div 2x^{3}$
7th-9th grade
Algebra
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search-thumbnail-Diketahui f(x)=x^{4}-8x^{2}+x-3 dan g(x)=x^{3}-2x^{2}+3x+4 maka nilai (fg)(x)= ..
Pertanyaan dan Pilihan Diketahui $f\left(x\right)=x^{4}-8x^{2}+x-3$ dan $g\left(x\right)=x^{3}-2x^{2}+3x+4$ maka nilai $\left(fg\right)\left(x\right)=$ .. $A$ A $x^{7}-10x^{6}-5x^{5}+21x^{4}-29x^{3}-23x^{2}-5x+12$ $B$ $x^{7}-10x^{6}-5x^{5}+21x^{4}-29x^{3}-23x^{2}+5x-12$ $c$ C $x^{7}-10x^{5}-5x^{5}+21x^{4}-29x^{3}-23x^{2}-5x-12$ $D$ D $x^{7}-10x^{6}-5x^{5}+21x^{4}+29x^{3}-23x^{2}-5x-12$ $E$ E $x^{7}-10x^{6}-5x^{5}+21x^{4}-29x^{3}+23x^{2}-5x-12$ total data
10th-13th grade
Other
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