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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
$y = x ^ { 2 } - 5 x + 6$
$y = 0$
$x$Intercept
$\left ( 3 , 0 \right )$, $\left ( 2 , 0 \right )$
$y$Intercept
$\left ( 0 , 6 \right )$
Minimum
$\left ( \dfrac { 5 } { 2 } , - \dfrac { 1 } { 4 } \right )$
Standard form
$y = \left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } - \dfrac { 1 } { 4 }$
$\begin{array} {l} x = 3 \\ x = 2 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = 2 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = - 6 + \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = - 6 + \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = 2 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 5 \right ) \pm \sqrt{ \left ( - 5 \right ) ^ { 2 } - 4 \times 1 \times 6 } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 5 \pm \sqrt{ \left ( - 5 \right ) ^ { 2 } - 4 \times 1 \times 6 } } { 2 \times 1 }$
$x = \dfrac { 5 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 6 } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 5 \pm \sqrt{ 5 ^ { 2 } - 4 \times 1 \times 6 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \pm \sqrt{ \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \pm \sqrt{ \color{#FF6800}{ 1 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$x = \dfrac { 5 \pm \sqrt{ \color{#FF6800}{ 1 } } } { 2 \times 1 }$
$n $ square root of 1 is 1 $ $
$x = \dfrac { 5 \pm \color{#FF6800}{ 1 } } { 2 \times 1 }$
$x = \dfrac { 5 \pm 1 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 5 \pm 1 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \pm \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } { 2 } \\ x = \dfrac { 5 - 1 } { 2 } \end{array}$
$ $ Add $ 5 $ and $ 1$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 6 } } { 2 } \\ x = \dfrac { 5 - 1 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 2 } } } \\ x = \dfrac { 5 - 1 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { 5 - 1 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { 5 - 1 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 3 } \\ x = \dfrac { 5 - 1 } { 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 2 } \end{array}$
$ $ Subtract $ 1 $ from $ 5$
$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ 4 } } { 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 1 } } } \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 1 } } } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ 2 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 6$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 5 ^ { 2 } - 4 \times 1 \times 6$
$D = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 6$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 25 } - 4 \times 1 \times 6$
$D = 25 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 6$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 25 - 4 \times 6$
$D = 25 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 }$
$ $ Multiply $ - 4 $ and $ 6$
$D = 25 \color{#FF6800}{ - } \color{#FF6800}{ 24 }$
$D = \color{#FF6800}{ 25 } \color{#FF6800}{ - } \color{#FF6800}{ 24 }$
$ $ Subtract $ 24 $ from $ 25$
$D = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 5 , \alpha \beta = 6$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { \color{#FF6800}{ 1 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 1 } } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { \color{#FF6800}{ 1 } } } , \alpha \beta = \dfrac { 6 } { 1 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 1 } } } , \alpha \beta = \dfrac { 6 } { 1 }$
$\alpha + \beta = \dfrac { 5 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 6 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = \color{#FF6800}{ 5 } , \alpha \beta = \dfrac { 6 } { 1 }$
$\alpha + \beta = 5 , \alpha \beta = \dfrac { 6 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 5 , \alpha \beta = \color{#FF6800}{ 6 }$
Solution search results
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search-thumbnail-t2 -5x 5x+6=0
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search-thumbnail-B =(x:x^{2}-5x+6=0,x∈N)
$B$ $=\left(x:x^{2}-5x+6=0,x∈N\right)$
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search-thumbnail-1 The equation whose roots are 1,3,-2 is
$1$ The equation whose roots are $1,3,-2$ is $1\right)$ $x^{3}+2x^{2}-5x+6=0$ $2\right)$ $x^{3}-2x^{4}+5x-6=0$ $31$ $x^{3}-2x^{2}-5x+6=0$ $4\right)$ $x^{1}-2x^{2}-5x-6=0$
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