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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
$y = x ^ { 2 }$
$y = \dfrac { \left ( x + 1 \right ) \left ( x + 2 \right ) } { 3 }$
$x$-intercept
$\left ( 0 , 0 \right )$
$y$-intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( 0 , 0 \right )$
Standard form
$y = x ^ { 2 }$
$x$-intercept
$\left ( - 1 , 0 \right )$, $\left ( - 2 , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 2 } { 3 } \right )$
Minimum
$\left ( - \dfrac { 3 } { 2 } , - \dfrac { 1 } { 12 } \right )$
Standard form
$y = \dfrac { 1 } { 3 } \left ( x + \dfrac { 3 } { 2 } \right ) ^ { 2 } - \dfrac { 1 } { 12 }$
$x ^{ 2 } = \dfrac{ \left( x+1 \right) \left( x+2 \right) }{ 3 }$
$\begin{array} {l} x = 2 \\ x = - \dfrac { 1 } { 2 } \end{array}$
Solve quadratic equations using the square root
$x ^ { 2 } = \color{#FF6800}{ \dfrac { \left ( x + 1 \right ) \left ( x + 2 \right ) } { 3 } }$
$ $ Arrange the fraction expression $ $
$x ^ { 2 } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$3 x ^ { 2 } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 2$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 3 x - 2 = 0$
$ $ Calculate between similar terms $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 3 x - 2 = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 3 } { 4 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - \dfrac { 3 } { 4 } \right ) ^ { 2 } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 3 } { 4 } \right ) ^ { 2 } = 1 + \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x - \dfrac { 3 } { 4 } \right ) ^ { 2 } = 1 + \dfrac { \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 ^ { 2 } } { 4 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 25 } { 16 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 25 } { 16 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 25 } { 16 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 25 } { 16 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 4 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 4 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{array}$
$\begin{array} {l} x = 2 \\ x = - \dfrac { 1 } { 2 } \end{array}$
Calculate using the quadratic formula
$x ^ { 2 } = \color{#FF6800}{ \dfrac { \left ( x + 1 \right ) \left ( x + 2 \right ) } { 3 } }$
$ $ Arrange the fraction expression $ $
$x ^ { 2 } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$3 x ^ { 2 } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 2$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 \pm \sqrt{ \left ( - 3 \right ) ^ { 2 } - 4 \times 2 \times \left ( - 2 \right ) } } { 2 \times 2 } }$
$x = \dfrac { 3 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 2 \right ) } } { 2 \times 2 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 3 \pm \sqrt{ 3 ^ { 2 } - 4 \times 2 \times \left ( - 2 \right ) } } { 2 \times 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 \pm \sqrt{ 3 ^ { 2 } - 4 \times 2 \times \left ( - 2 \right ) } } { 2 \times 2 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 \pm \sqrt{ 25 } } { 2 \times 2 } }$
$x = \dfrac { 3 \pm \sqrt{ \color{#FF6800}{ 25 } } } { 2 \times 2 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 3 \pm \color{#FF6800}{ 5 } } { 2 \times 2 }$
$x = \dfrac { 3 \pm 5 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$ $ Multiply $ 2 $ and $ 2$
$x = \dfrac { 3 \pm 5 } { \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 \pm 5 } { 4 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 + 5 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 - 5 } { 4 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } } { 4 } \\ x = \dfrac { 3 - 5 } { 4 } \end{array}$
$ $ Add $ 3 $ and $ 5$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 8 } } { 4 } \\ x = \dfrac { 3 - 5 } { 4 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 8 } { 4 } } \\ x = \dfrac { 3 - 5 } { 4 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 } { 1 } } \\ x = \dfrac { 3 - 5 } { 4 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 } { 1 } } \\ x = \dfrac { 3 - 5 } { 4 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 2 } \\ x = \dfrac { 3 - 5 } { 4 } \end{array}$
$\begin{array} {l} x = 2 \\ x = \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { 4 } \end{array}$
$ $ Subtract $ 5 $ from $ 3$
$\begin{array} {l} x = 2 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 4 } \end{array}$
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ \dfrac { - 2 } { 4 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ \dfrac { - 1 } { 2 } } \end{array}$
$\begin{array} {l} x = 2 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 2 } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$x ^ { 2 } = \color{#FF6800}{ \dfrac { \left ( x + 1 \right ) \left ( x + 2 \right ) } { 3 } }$
$ $ Arrange the fraction expression $ $
$x ^ { 2 } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$3 x ^ { 2 } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 2$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 3 x - 2 = 0$
$ $ Calculate between similar terms $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 3 x - 2 = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 2 \right )$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 3 ^ { 2 } - 4 \times 2 \times \left ( - 2 \right )$
$D = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 2 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 9 } - 4 \times 2 \times \left ( - 2 \right )$
$D = 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$ $ Multiply the numbers $ $
$D = 9 + \color{#FF6800}{ 16 }$
$D = \color{#FF6800}{ 9 } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
$ $ Add $ 9 $ and $ 16$
$D = \color{#FF6800}{ 25 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 25 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = \dfrac { 3 } { 2 } , \alpha \beta = - 1$
Find the sum and product of the two roots of the quadratic equation
$x ^ { 2 } = \color{#FF6800}{ \dfrac { \left ( x + 1 \right ) \left ( x + 2 \right ) } { 3 } }$
$ $ Arrange the fraction expression $ $
$x ^ { 2 } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { x ^ { 2 } + 3 x + 2 } { 3 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$3 x ^ { 2 } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 2$
$ $ Move the expression to the left side and change the symbol $ $
$3 x ^ { 2 } - x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 3 x - 2 = 0$
$ $ Calculate between similar terms $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 3 x - 2 = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 3 } { 2 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 2 } { 2 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 3 } { 2 } } , \alpha \beta = \dfrac { - 2 } { 2 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 3 } { 2 } } , \alpha \beta = \dfrac { - 2 } { 2 }$
$\alpha + \beta = \dfrac { 3 } { 2 } , \alpha \beta = \color{#FF6800}{ \dfrac { - 2 } { 2 } }$
$ $ Reduce the fraction $ $
$\alpha + \beta = \dfrac { 3 } { 2 } , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ 그래프 보기 $ $
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search-thumbnail-If the sum of two consecutive numbers is $45$ and one number is $X$ .This statement in the form of equation $1s:$ $\left(1$ Point) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$
7th-9th grade
Algebra
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search-thumbnail-In the following problem $a,b,$ b,and c represent REAL NUMBERS. The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ $○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ $○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) $○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ $○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ $○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ $○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$
Calculus
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search-thumbnail-The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
Calculus
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search-thumbnail- $\right)c$ $c$ $co$ $3c3x$ $x$ Question 2 $1pts$ For each of the following questions, clearly select the ONE correct answer. Which of the following functions is increasing on the interval $\left(0,1\right)7$ $Ap$ $Ass$ Coe $-lnlef\left(x\right)ngh$ Col $x-ln\left(x\right)$ $x-1$ $Co$ $○y$ $xcos\left(2x\right)$ Diff Eva $O-\dfrac {3x} {x-1}-1trac\left(3x\right)\left(x-1\right)$ Fac Inte $○x$ $x$ $ln\left(x\right)$ Lim Seri Sim Solu $Ne\times t$ > Con $Con$ $Con$
Calculus
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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