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Solve the quadratic equation
Answer
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Number of solution
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Relationship between roots and coefficients
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Graph
$y = \left ( x - 2 \right ) \left ( x + 3 \right )$
$y = 5$
$x$Intercept
$\left ( 2 , 0 \right )$, $\left ( - 3 , 0 \right )$
$y$Intercept
$\left ( 0 , - 6 \right )$
Minimum
$\left ( - \dfrac { 1 } { 2 } , - \dfrac { 25 } { 4 } \right )$
Standard form
$y = \left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } - \dfrac { 25 } { 4 }$
$\begin{array} {l} x = \dfrac { - 1 + 3 \sqrt{ 5 } } { 2 } \\ x = \dfrac { - 1 - 3 \sqrt{ 5 } } { 2 } \end{array}$
Solve quadratic equations using the square root
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 5$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = 5$
$x ^ { 2 } + x - 6 = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + x - 6 \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$ $ Find the sum of the negative numbers $ $
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 11 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 11 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ 11 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = 11 + \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = 11 + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 11 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 45 } } { \color{#FF6800}{ 4 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 45 } } { \color{#FF6800}{ 4 } } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 45 } } { \color{#FF6800}{ 4 } } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 45 } } { \color{#FF6800}{ 4 } } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} x = \dfrac { - 1 + 3 \sqrt{ 5 } } { 2 } \\ x = \dfrac { - 1 - 3 \sqrt{ 5 } } { 2 } \end{array}$
Calculate using the quadratic formula
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 5$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = 5$
$x ^ { 2 } + x - 6 = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + x - 6 \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$ $ Find the sum of the negative numbers $ $
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 11 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 11 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \pm \sqrt{ \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \pm \sqrt{ \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \pm \sqrt{ \color{#FF6800}{ 45 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$x = \dfrac { - 1 \pm \sqrt{ \color{#FF6800}{ 45 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 1 \pm \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { 2 \times 1 }$
$x = \dfrac { - 1 \pm 3 \sqrt{ 5 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 1 \pm 3 \sqrt{ 5 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \pm \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 5 } } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 5$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = 5$
$x ^ { 2 } + x - 6 = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + x - 6 \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$ $ Find the sum of the negative numbers $ $
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 11 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 11 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right )$
$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 11 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1 } - 4 \times 1 \times \left ( - 11 \right )$
$D = 1 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 11 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 1 - 4 \times \left ( - 11 \right )$
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right )$
$ $ Multiply $ - 4 $ and $ - 11$
$D = 1 + \color{#FF6800}{ 44 }$
$D = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 44 }$
$ $ Add $ 1 $ and $ 44$
$D = \color{#FF6800}{ 45 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 45 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 1 , \alpha \beta = - 11$
Find the sum and product of the two roots of the quadratic equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) = 5$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = 5$
$x ^ { 2 } + x - 6 = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$x ^ { 2 } + x - 6 \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$ $ Find the sum of the negative numbers $ $
$x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 11 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 11 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 1 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { \color{#FF6800}{ 1 } } }$
$\alpha + \beta = - \dfrac { 1 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 11 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 1 } , \alpha \beta = \dfrac { - 11 } { 1 }$
$\alpha + \beta = - 1 , \alpha \beta = \dfrac { - 11 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 1 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 11 }$
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