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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
$y = 3 x ^ { 2 } + 3 x + 1$
$y = 0$
$y$Intercept
$\left ( 0 , 1 \right )$
Minimum
$\left ( - \dfrac { 1 } { 2 } , \dfrac { 1 } { 4 } \right )$
Standard form
$y = 3 \left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } + \dfrac { 1 } { 4 }$
$3x ^{ 2 } +3x+1 = 0$
$\begin{array} {l} x = \dfrac { - 3 + \sqrt{ 3 } i } { 6 } \\ x = \dfrac { - 3 - \sqrt{ 3 } i } { 6 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = - \dfrac { 1 } { 3 } + \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = - \dfrac { 1 } { 3 } + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 ^ { 2 } } { 2 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 3 } i } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 3 } i } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 3 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 3 } i } { 6 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 3 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 3 } i } { 6 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 + \sqrt{ 3 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 - \sqrt{ 3 } i } { 6 } } \end{array}$
$\begin{array} {l} x = \dfrac { - 3 + \sqrt{ 3 } i } { 6 } \\ x = \dfrac { - 3 - \sqrt{ 3 } i } { 6 } \end{array}$
Calculate using the quodratic formula$($Imaginary root solution$)
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ 3 ^ { 2 } - 4 \times 3 \times 1 } } { 2 \times 3 } }$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 1 } } { 2 \times 3 }$
$ $ Calculate power $ $
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 9 } - 4 \times 3 \times 1 } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ 9 - 4 \times 3 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } } { 2 \times 3 }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 3 \pm \sqrt{ 9 - 4 \times 3 } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } } } { 2 \times 3 }$
$ $ Multiply $ - 4 $ and $ 3$
$x = \dfrac { - 3 \pm \sqrt{ 9 \color{#FF6800}{ - } \color{#FF6800}{ 12 } } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 12 } } } { 2 \times 3 }$
$ $ Subtract $ 12 $ from $ 9$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 3 } } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ - 3 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { - 3 \pm \sqrt{ - 3 } } { \color{#FF6800}{ 6 } }$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ - } 3 } } { 6 }$
$ $ Subtracting (-) from the square root gives i $ $
$x = \dfrac { - 3 \pm \sqrt{ 3 } \color{#FF6800}{ i } } { 6 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ 3 } i } { 6 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 + \sqrt{ 3 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 - \sqrt{ 3 } i } { 6 } } \end{array}$
$ $ Do not have the solution $ $
Calculate using the quadratic formula
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 1 } } { 2 \times 3 }$
$ $ Calculate power $ $
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 9 } - 4 \times 3 \times 1 } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ 9 - 4 \times 3 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } } { 2 \times 3 }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 3 \pm \sqrt{ 9 - 4 \times 3 } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } } } { 2 \times 3 }$
$ $ Multiply $ - 4 $ and $ 3$
$x = \dfrac { - 3 \pm \sqrt{ 9 \color{#FF6800}{ - } \color{#FF6800}{ 12 } } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 12 } } } { 2 \times 3 }$
$ $ Subtract $ 12 $ from $ 9$
$x = \dfrac { - 3 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 3 } } } { 2 \times 3 }$
$x = \dfrac { - 3 \pm \sqrt{ - 3 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { - 3 \pm \sqrt{ - 3 } } { \color{#FF6800}{ 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 3 \pm \sqrt{ - 3 } } { 6 } }$
$ $ The square root of a negative number does not exist within the set of real numbers $ $
$ $ Do not have the solution $ $
$ $ Do not have the real root $ $
Find the number of solutions
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
$D = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 1$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 9 } - 4 \times 3 \times 1$
$D = 9 - 4 \times 3 \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 9 - 4 \times 3$
$D = 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 }$
$ $ Multiply $ - 4 $ and $ 3$
$D = 9 \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$D = \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$ $ Subtract $ 12 $ from $ 9$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$ $ Since $ D<0 $ , there is no real root of the following quadratic equation $ $
$ $ Do not have the real root $ $
$\alpha + \beta = - 1 , \alpha \beta = \dfrac { 1 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 3 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
$\alpha + \beta = - \color{#FF6800}{ \dfrac { 3 } { 3 } } , \alpha \beta = \dfrac { 1 } { 3 }$
$ $ Reduce the fraction $ $
$\alpha + \beta = - \color{#FF6800}{ 1 } , \alpha \beta = \dfrac { 1 } { 3 }$
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