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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
$y = x ^ { 2 } + x - 182$
$y = 0$
$x$Intercept
$\left ( - 14 , 0 \right )$, $\left ( 13 , 0 \right )$
$y$Intercept
$\left ( 0 , - 182 \right )$
Minimum
$\left ( - \dfrac { 1 } { 2 } , - \dfrac { 729 } { 4 } \right )$
Standard form
$y = \left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } - \dfrac { 729 } { 4 }$
$x ^{ 2 } +x-182 = 0$
$\begin{array} {l} x = 13 \\ x = - 14 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 182 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 13 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 14 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 13 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 14 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 13 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 14 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 13 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 14 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 13 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 14 } \end{array}$
$\begin{array} {l} x = 13 \\ x = - 14 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 182 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 182 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 182 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ 182 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = 182 + \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 1 } { 2 } \right ) ^ { 2 } = 182 + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 182 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 ^ { 2 } } { 2 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 729 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 729 } { 4 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 729 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 729 } { 4 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 27 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 27 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 27 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 27 } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 27 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 27 } { 2 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 13 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 14 } \end{array}$
$\begin{array} {l} x = 13 \\ x = - 14 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 \pm \sqrt{ 1 ^ { 2 } - 4 \times 1 \times \left ( - 182 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 \pm \sqrt{ 729 } } { 2 \times 1 } }$
$x = \dfrac { - 1 \pm \sqrt{ \color{#FF6800}{ 729 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 1 \pm \color{#FF6800}{ 27 } } { 2 \times 1 }$
$x = \dfrac { - 1 \pm 27 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 1 \pm 27 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 \pm 27 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 + 27 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 - 27 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 27 } } { 2 } \\ x = \dfrac { - 1 - 27 } { 2 } \end{array}$
$ $ Add $ - 1 $ and $ 27$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 26 } } { 2 } \\ x = \dfrac { - 1 - 27 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 26 } { 2 } } \\ x = \dfrac { - 1 - 27 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 13 } { 1 } } \\ x = \dfrac { - 1 - 27 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 13 } { 1 } } \\ x = \dfrac { - 1 - 27 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 13 } \\ x = \dfrac { - 1 - 27 } { 2 } \end{array}$
$\begin{array} {l} x = 13 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 27 } } { 2 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 13 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 28 } } { 2 } \end{array}$
$\begin{array} {l} x = 13 \\ x = \color{#FF6800}{ \dfrac { - 28 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 13 \\ x = \color{#FF6800}{ \dfrac { - 14 } { 1 } } \end{array}$
$\begin{array} {l} x = 13 \\ x = \dfrac { - 14 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 13 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 14 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 182 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 182 } \right )$
$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 182 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1 } - 4 \times 1 \times \left ( - 182 \right )$
$D = 1 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 182 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 1 - 4 \times \left ( - 182 \right )$
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 182 } \right )$
$ $ Multiply $ - 4 $ and $ - 182$
$D = 1 + \color{#FF6800}{ 728 }$
$D = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 728 }$
$ $ Add $ 1 $ and $ 728$
$D = \color{#FF6800}{ 729 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 729 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 1 , \alpha \beta = - 182$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 182 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 182 } { 1 } }$
$\alpha + \beta = - \dfrac { 1 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 182 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 1 } , \alpha \beta = \dfrac { - 182 } { 1 }$
$\alpha + \beta = - 1 , \alpha \beta = \dfrac { - 182 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 1 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 182 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-e. $C=\left(x:x^{2}+x-12=0x∈N$
7th-9th grade
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search-thumbnail-$x^{2}+x-8=0$ $x^{2}-7x-8=0$ 
$O$ $O$ 
$x^{2+5x-14}=0$ $x^{2}-5x-14=0$
7th-9th grade
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search-thumbnail-$20.$ Find the equations of straight lines passing through $\left(-2-7$ and having an intercept of length $3$ 
between the straight lines $4x+3y=12$ and $4x+3y=3$ are 
$x-2=0.7x-24y+182=0$ 
$x-3=0.7x-24y+182=0$ 
$x-2=0.7x-2y+182=0$ 
$x+2=0.7x+24y+182=0$
10th-13th grade
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