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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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$y = x ^ { 2 } + \left ( x + 1 \right ) ^ { 2 }$
$y = 113$
$y$Intercept
$\left ( 0 , 1 \right )$
Minimum
$\left ( - \dfrac { 1 } { 2 } , \dfrac { 1 } { 2 } \right )$
Standard form
$y = 2 \left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } + \dfrac { 1 } { 2 }$
$\begin{array} {l} x = 7 \\ x = - 8 \end{array}$
Find solution by method of factorization
$x ^ { 2 } + \left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 113 }$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + \left ( x + 1 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$ $ Expand the expression $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right ) = 0$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$
$\begin{array} {l} x = 7 \\ x = - 8 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = 113$
$ $ Organize the expression $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 113$
$2 x ^ { 2 } + 2 x + 1 = \color{#FF6800}{ 113 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + 2 x + 1 \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$2 x ^ { 2 } + 2 x + \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$ $ Subtract $ 113 $ from $ 1$
$2 x ^ { 2 } + 2 x \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 56 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 56 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 56 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 56 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ 56 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = 56 + \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = 56 + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 56 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 225 } } { \color{#FF6800}{ 4 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 225 } } { \color{#FF6800}{ 4 } } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 225 } } { \color{#FF6800}{ 4 } } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 225 } } { \color{#FF6800}{ 4 } } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 7 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$
$\begin{array} {l} x = 7 \\ x = - 8 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = 113$
$ $ Organize the expression $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 113$
$2 x ^ { 2 } + 2 x + 1 = \color{#FF6800}{ 113 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + 2 x + 1 \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$2 x ^ { 2 } + 2 x + \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$ $ Subtract $ 113 $ from $ 1$
$2 x ^ { 2 } + 2 x \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 112 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 112 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \pm \sqrt{ \color{#FF6800}{ 900 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } } }$
$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ 900 } } } { 2 \times 2 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 2 \pm \color{#FF6800}{ 30 } } { 2 \times 2 }$
$x = \dfrac { - 2 \pm 30 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$ $ Multiply $ 2 $ and $ 2$
$x = \dfrac { - 2 \pm 30 } { \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \pm \color{#FF6800}{ 30 } } { \color{#FF6800}{ 4 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 30 } } { \color{#FF6800}{ 4 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 30 } } { \color{#FF6800}{ 4 } } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 30 } } { 4 } \\ x = \dfrac { - 2 - 30 } { 4 } \end{array}$
$ $ Add $ - 2 $ and $ 30$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 28 } } { 4 } \\ x = \dfrac { - 2 - 30 } { 4 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 28 } } { \color{#FF6800}{ 4 } } } \\ x = \dfrac { - 2 - 30 } { 4 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { - 2 - 30 } { 4 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { - 2 - 30 } { 4 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 7 } \\ x = \dfrac { - 2 - 30 } { 4 } \end{array}$
$\begin{array} {l} x = 7 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 30 } } { 4 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 7 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 32 } } { 4 } \end{array}$
$\begin{array} {l} x = 7 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 32 } } { \color{#FF6800}{ 4 } } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 7 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } } { \color{#FF6800}{ 1 } } } \end{array}$
$\begin{array} {l} x = 7 \\ x = \dfrac { - 8 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 7 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = 113$
$ $ Organize the expression $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 113$
$2 x ^ { 2 } + 2 x + 1 = \color{#FF6800}{ 113 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + 2 x + 1 \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$2 x ^ { 2 } + 2 x + \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$ $ Subtract $ 113 $ from $ 1$
$2 x ^ { 2 } + 2 x \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 112 } \right )$
$D = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 112 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 4 } - 4 \times 2 \times \left ( - 112 \right )$
$D = 4 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 112 } \right )$
$ $ Multiply the numbers $ $
$D = 4 + \color{#FF6800}{ 896 }$
$D = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 896 }$
$ $ Add $ 4 $ and $ 896$
$D = \color{#FF6800}{ 900 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 900 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 1 , \alpha \beta = - 56$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = 113$
$ $ Organize the expression $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 113$
$2 x ^ { 2 } + 2 x + 1 = \color{#FF6800}{ 113 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + 2 x + 1 \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$2 x ^ { 2 } + 2 x + \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 113 } = 0$
$ $ Subtract $ 113 $ from $ 1$
$2 x ^ { 2 } + 2 x \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 2 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 112 } } { \color{#FF6800}{ 2 } } }$
$\alpha + \beta = - \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 2 } } } , \alpha \beta = \dfrac { - 112 } { 2 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = - \color{#FF6800}{ 1 } , \alpha \beta = \dfrac { - 112 } { 2 }$
$\alpha + \beta = - 1 , \alpha \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 112 } } { \color{#FF6800}{ 2 } } }$
$ $ Reduce the fraction $ $
$\alpha + \beta = - 1 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 56 }$
Solution search results
search-thumbnail-If the sum of two consecutive
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-Theis the statement of the Mean Value Theorem from your te\timestb00k
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
Search count: 3,278
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search-thumbnail-s|ef(-1n }\right)^{50}\ ) \ is\ equal\ to\ S
$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th grade
Other
Search count: 625
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search-thumbnail-$The\ reciprocal\\ 0+11) \left(\frac{2}
$8$ $\left(1$ Point) $1\right)$ The\ reciprocal\\ $0+11\right)$ \left(\frac{2} $c\left(2\right)$ {5}\right)^0\ $\right)$ \ $1111s\right)$ $S$ $S1S$ $s3S$ $S4S$ $s2S$
7th-9th grade
Other
Search count: 4,895
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search-thumbnail-Given the set of ordered pairs ((-7.0),(-6,5),(-5,-3),(-1,2) (1,6),(2,-2) (5,3)(7,-8))
Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ Find f(7)fAleft(7\right) O a O b -8 6. $5$
7th-9th grade
Algebra
Search count: 119
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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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