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Formula

Solve the quadratic equation

Answer

Number of solution

Answer

Relationship between roots and coefficients

Answer

Graph

$y = 2 x ^ { 2 } - x - 15$

$y = 0$

$x$-intercept

$\left ( 3 , 0 \right )$, $\left ( - \dfrac { 5 } { 2 } , 0 \right )$

$y$-intercept

$\left ( 0 , - 15 \right )$

Minimum

$\left ( \dfrac { 1 } { 4 } , - \dfrac { 121 } { 8 } \right )$

Standard form

$y = 2 \left ( x - \dfrac { 1 } { 4 } \right ) ^ { 2 } - \dfrac { 121 } { 8 }$

$2x ^{ 2 } -x-15 = 0$

$\begin{array} {l} x = 3 \\ x = - \dfrac { 5 } { 2 } \end{array}$

Find solution by method of factorization

$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$

$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = 0$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 0 }$

$ $ If the product of the factor is 0, at least one factor should be 0 $ $

$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \end{array}$

$ $ Solve the equation to find $ x$

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{array}$

$\begin{array} {l} x = 3 \\ x = - \dfrac { 5 } { 2 } \end{array}$

Solve quadratic equations using the square root

$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$

$ $ Divide both sides by the coefficient of the leading highest term $ $

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } = \color{#FF6800}{ 0 }$

$ $ Convert the quadratic expression on the left side to a perfect square format $ $

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$

$\left ( x - \dfrac { 1 } { 4 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$

$ $ Move the constant to the right side and change the sign $ $

$\left ( x - \dfrac { 1 } { 4 } \right ) ^ { 2 } = \color{#FF6800}{ \dfrac { 15 } { 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$

$\left ( x - \dfrac { 1 } { 4 } \right ) ^ { 2 } = \dfrac { 15 } { 2 } + \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$

$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $

$\left ( x - \dfrac { 1 } { 4 } \right ) ^ { 2 } = \dfrac { 15 } { 2 } + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } }$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 15 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 ^ { 2 } } { 4 ^ { 2 } } }$

$ $ Organize the expression $ $

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 121 } { 16 } }$

$ $ Solve quadratic equations using the square root $ $

$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 121 } { 16 } } }$

$ $ Solve a solution to $ x$

$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 11 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 4 } } \end{array}$

$ $ Organize the expression $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{array}$

$\begin{array} {l} x = 3 \\ x = - \dfrac { 5 } { 2 } \end{array}$

Calculate using the quadratic formula

$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 1 \right ) \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 2 \times \left ( - 15 \right ) } } { 2 \times 2 }$

$ $ Simplify Minus $ $

$x = \dfrac { 1 \pm \sqrt{ \left ( - 1 \right ) ^ { 2 } - 4 \times 2 \times \left ( - 15 \right ) } } { 2 \times 2 }$

$x = \dfrac { 1 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 15 \right ) } } { 2 \times 2 }$

$ $ Remove negative signs because negative numbers raised to even powers are positive $ $

$x = \dfrac { 1 \pm \sqrt{ 1 ^ { 2 } - 4 \times 2 \times \left ( - 15 \right ) } } { 2 \times 2 }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 \pm \sqrt{ 1 ^ { 2 } - 4 \times 2 \times \left ( - 15 \right ) } } { 2 \times 2 } }$

$ $ Organize the expression $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 \pm \sqrt{ 121 } } { 2 \times 2 } }$

$x = \dfrac { 1 \pm \sqrt{ \color{#FF6800}{ 121 } } } { 2 \times 2 }$

$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $

$x = \dfrac { 1 \pm \color{#FF6800}{ 11 } } { 2 \times 2 }$

$x = \dfrac { 1 \pm 11 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$

$ $ Multiply $ 2 $ and $ 2$

$x = \dfrac { 1 \pm 11 } { \color{#FF6800}{ 4 } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 \pm 11 } { 4 } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 + 11 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 - 11 } { 4 } } \end{array}$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 11 } } { 4 } \\ x = \dfrac { 1 - 11 } { 4 } \end{array}$

$ $ Add $ 1 $ and $ 11$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 12 } } { 4 } \\ x = \dfrac { 1 - 11 } { 4 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 12 } { 4 } } \\ x = \dfrac { 1 - 11 } { 4 } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \\ x = \dfrac { 1 - 11 } { 4 } \end{array}$

$ $ Reduce the fraction to the lowest term $ $

$\begin{array} {l} x = \color{#FF6800}{ 3 } \\ x = \dfrac { 1 - 11 } { 4 } \end{array}$

$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { 4 } \end{array}$

$ $ Subtract $ 11 $ from $ 1$

$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 4 } \end{array}$

$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { - 10 } { 4 } } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { - 5 } { 2 } } \end{array}$

$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { 2 } \end{array}$

$ $ Move the minus sign to the front of the fraction $ $

$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{array}$

$ $ 2 real roots $ $

Find the number of solutions

$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$

$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$

$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right )$

$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 15 \right )$

$ $ Remove negative signs because negative numbers raised to even powers are positive $ $

$D = 1 ^ { 2 } - 4 \times 2 \times \left ( - 15 \right )$

$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 15 \right )$

$ $ Calculate power $ $

$D = \color{#FF6800}{ 1 } - 4 \times 2 \times \left ( - 15 \right )$

$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right )$

$ $ Multiply the numbers $ $

$D = 1 + \color{#FF6800}{ 120 }$

$D = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 120 }$

$ $ Add $ 1 $ and $ 120$

$D = \color{#FF6800}{ 121 }$

$\color{#FF6800}{ D } = \color{#FF6800}{ 121 }$

$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $

$ $ 2 real roots $ $

$\alpha + \beta = \dfrac { 1 } { 2 } , \alpha \beta = - \dfrac { 15 } { 2 }$

Find the sum and product of the two roots of the quadratic equation

$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$

$ $ 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}$

$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 2 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 15 } { 2 } }$

$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 2 } } , \alpha \beta = \dfrac { - 15 } { 2 }$

$ $ Solve the sign of a fraction with a negative sign $ $

$\alpha + \beta = \color{#FF6800}{ \dfrac { 1 } { 2 } } , \alpha \beta = \dfrac { - 15 } { 2 }$

$\alpha + \beta = \dfrac { 1 } { 2 } , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { 2 }$

$ $ Move the minus sign to the front of the fraction $ $

$\alpha + \beta = \dfrac { 1 } { 2 } , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$

$ $ 그래프 보기 $ $

Graph

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