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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
$y = 3 x ^ { 2 } - x + 2$
$y = 0$
$y$Intercept
$\left ( 0 , 2 \right )$
Minimum
$\left ( \dfrac { 1 } { 6 } , \dfrac { 23 } { 12 } \right )$
Standard form
$y = 3 \left ( x - \dfrac { 1 } { 6 } \right ) ^ { 2 } + \dfrac { 23 } { 12 }$
$3x ^{ 2 } -x+2 = 0$
$\begin{array} {l} x = \dfrac { 1 + \sqrt{ 23 } i } { 6 } \\ x = \dfrac { 1 - \sqrt{ 23 } i } { 6 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 1 } { 6 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - \dfrac { 1 } { 6 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 1 } { 6 } \right ) ^ { 2 } = - \dfrac { 2 } { 3 } + \left ( \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x - \dfrac { 1 } { 6 } \right ) ^ { 2 } = - \dfrac { 2 } { 3 } + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 ^ { 2 } } { 6 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 23 } { 36 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 23 } { 36 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 23 } { 36 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 23 } { 36 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 23 } i } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 23 } i } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 23 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 23 } i } { 6 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 23 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 23 } i } { 6 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 + \sqrt{ 23 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 - \sqrt{ 23 } i } { 6 } } \end{array}$
$\begin{array} {l} x = \dfrac { 1 + \sqrt{ 23 } i } { 6 } \\ x = \dfrac { 1 - \sqrt{ 23 } i } { 6 } \end{array}$
Calculate using the quodratic formula$($Imaginary root solution$)
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - \left ( - 1 \right ) \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 } }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 1 \right ) \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$ $ Simplify Minus $ $
$x = \dfrac { 1 \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 1 \pm \sqrt{ 1 ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$ $ Calculate power $ $
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 1 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } } } { 2 \times 3 }$
$ $ Multiply the numbers $ $
$x = \dfrac { 1 \pm \sqrt{ 1 \color{#FF6800}{ - } \color{#FF6800}{ 24 } } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 24 } } } { 2 \times 3 }$
$ $ Subtract $ 24 $ from $ 1$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 23 } } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ - 23 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { 1 \pm \sqrt{ - 23 } } { \color{#FF6800}{ 6 } }$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ - } 23 } } { 6 }$
$ $ Subtracting (-) from the square root gives i $ $
$x = \dfrac { 1 \pm \sqrt{ 23 } \color{#FF6800}{ i } } { 6 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 \pm \sqrt{ 23 } i } { 6 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 + \sqrt{ 23 } i } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 - \sqrt{ 23 } i } { 6 } } \end{array}$
$ $ Do not have the solution $ $
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 1 \right ) \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$ $ Simplify Minus $ $
$x = \dfrac { 1 \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 1 \pm \sqrt{ 1 ^ { 2 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$ $ Calculate power $ $
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 1 } - 4 \times 3 \times 2 } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } } } { 2 \times 3 }$
$ $ Multiply the numbers $ $
$x = \dfrac { 1 \pm \sqrt{ 1 \color{#FF6800}{ - } \color{#FF6800}{ 24 } } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 24 } } } { 2 \times 3 }$
$ $ Subtract $ 24 $ from $ 1$
$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 23 } } } { 2 \times 3 }$
$x = \dfrac { 1 \pm \sqrt{ - 23 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
$ $ Multiply $ 2 $ and $ 3$
$x = \dfrac { 1 \pm \sqrt{ - 23 } } { \color{#FF6800}{ 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 \pm \sqrt{ - 23 } } { 6 } }$
$ $ 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
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 2$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 1 ^ { 2 } - 4 \times 3 \times 2$
$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 2$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1 } - 4 \times 3 \times 2$
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 }$
$ $ Multiply the numbers $ $
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 24 }$
$D = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 24 }$
$ $ Subtract $ 24 $ from $ 1$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 23 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ - } \color{#FF6800}{ 23 }$
$ $ Since $ D<0 $ , there is no real root of the following quadratic equation $ $
$ $ Do not have the real root $ $
$\alpha + \beta = \dfrac { 1 } { 3 } , \alpha \beta = \dfrac { 2 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 3 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 2 } { 3 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 3 } } , \alpha \beta = \dfrac { 2 } { 3 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 1 } { 3 } } , \alpha \beta = \dfrac { 2 } { 3 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$Q.11\right)$ If each root of the equation $3x^{2}-7x+4=0$ is increased by $y2,$ then the resulting equation is 
$3x^{2}-19x+30=0$ 
$○$ $3x^{2}+5x+2=0$ 
$○$ $3x^{2}-19x+2=0$ 
$3x^{2}-19x+20=0$
10th-13th grade
Trigonometry
search-thumbnail-$-5x^{2}-2+5x=$ 
$3x^{2}-5x+2=0$ 
$3x^{2}-3x-2x+2$ $3x$ $\left(x-1\right)^{-2}\left(x-1$ 
$x=\dfrac {2} {3}$ $x=4$
10th-13th grade
Other
search-thumbnail-
$1.$ $x^{2+3}=0$ $2.$ $2x^{2}+x+1=0$ $3$ $x^{2}+3x+9=\tarc{θ} $ 
4. $-x^{2}+x-2=0$ $5.$ $x^{2}+3x+5=0$ 6. $x^{2}-x+2=0$ 
$D$ $\sqrt{2} x^{2}+x+\sqrt{2} =0$ $\right)$ ) $\sqrt{3} x^{2}-\sqrt{2} x+3\sqrt{3} =0$ 
$9.$ $x^{2}+x+\dfrac {1} {\sqrt{2} }=0$ $\left(0\right)$ $x^{2}+\dfrac {x} {\sqrt{2} }+1=0$
10th-13th grade
Other
search-thumbnail-If $α,β,y$ are the roots of a cubic equation satisfying the relations 
$α+β+γ=2.α^{2}+β^{2}+γ^{2}=6$ and $α^{3}+β^{3}+γ^{3}=8$ then the equation $1s→$ 
$A\right)$ $x^{3}+2x^{2}-x+2=0$ $B\right)$ $x^{3}-2x^{2}-x+2=0$ 
$\left(\right)$ $x^{3}-2x^{2}+x+2=0$ D) $\right)$ $x^{3}-3x^{2}-x+2=0$
10th-13th grade
Algebra
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