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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
$y = x ^ { 2 } + 4$
$y = 0$
$y$-intercept
$\left ( 0 , 4 \right )$
Minimum
$\left ( 0 , 4 \right )$
Standard form
$y = x ^ { 2 } + 4$
$\left( x ^{ 2 } +4 \right) = 0$
$\begin{array} {l} x = 2 i \\ x = - 2 i \end{array}$
Solve quadratic equations using the square root
$x ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = 0$
$ $ Move the constant to the right side and change the sign $ $
$x ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 4 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 2 } \color{#FF6800}{ i }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 2 } \color{#FF6800}{ i }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \color{#FF6800}{ i } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ i } \end{array}$
$\begin{array} {l} x = 2 i \\ x = - 2 i \end{array}$
Calculate using the quodratic formula$($Imaginary root solution$)
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 0 \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times 4 } } { 2 \times 1 } }$
$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$ $ 0 has no sign $ $
$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$ $ The power of 0 is 0 $ $
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$ $ 0 does not change when you add or subtract $ $
$x = \dfrac { 0 \pm \sqrt{ - 4 \times 1 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 0 \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } { 2 \times 1 }$
$x = \dfrac { 0 \pm 4 \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } { 2 \times 1 }$
$ $ It is $ \sqrt{-1} = i$
$x = \dfrac { 0 \pm 4 \color{#FF6800}{ i } } { 2 \times 1 }$
$x = \dfrac { 0 \pm 4 i } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 0 \pm 4 i } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm 4 i } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 4 i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 4 i } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 4 i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 4 i } { 2 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 4 i } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { 4 i } { 2 } \\ x = \color{#FF6800}{ \dfrac { - 4 i } { 2 } } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = \dfrac { 4 i } { 2 } \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 i } { 2 } } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 4 i } { 2 } } \\ x = - \dfrac { 4 i } { 2 } \end{array}$
$ $ Reduce the fraction $ $
$\begin{array} {l} x = \color{#FF6800}{ 2 } \color{#FF6800}{ i } \\ x = - \dfrac { 4 i } { 2 } \end{array}$
$\begin{array} {l} x = 2 i \\ x = - \color{#FF6800}{ \dfrac { 4 i } { 2 } } \end{array}$
$ $ Reduce the fraction $ $
$\begin{array} {l} x = 2 i \\ x = - \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ i } \right ) \end{array}$
$\begin{array} {l} x = 2 i \\ x = \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ i } \right ) \end{array}$
$ $ Get rid of unnecessary parentheses $ $
$\begin{array} {l} x = 2 i \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ i } \end{array}$
$ $ Do not have the solution $ $
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$ $ 0 has no sign $ $
$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$ $ The power of 0 is 0 $ $
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times 4 } } { 2 \times 1 }$
$ $ 0 does not change when you add or subtract $ $
$x = \dfrac { 0 \pm \sqrt{ - 4 \times 1 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 4 } } { 2 \times 1 }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 0 \pm \sqrt{ - 4 \times 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } } } { 2 \times 1 }$
$ $ Multiply $ - 4 $ and $ 4$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 16 } } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ - 16 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 0 \pm \sqrt{ - 16 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ - 16 } } { 2 } }$
$ $ 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
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 }$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 4$
$ $ The power of 0 is 0 $ $
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times 4$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times 4$
$ $ 0 does not change when you add or subtract $ $
$D = - 4 \times 1 \times 4$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 4$
$ $ Multiplying any number by 1 does not change the value $ $
$D = - 4 \times 4$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 }$
$ $ Multiply $ - 4 $ and $ 4$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 16 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ - } \color{#FF6800}{ 16 }$
$ $ Since $ D<0 $ , there is no real root of the following quadratic equation $ $
$ $ Do not have the real root $ $
$\alpha + \beta = 0 , \alpha \beta = 4$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 4 } { 1 } }$
$\alpha + \beta = - \dfrac { 0 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 4 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { 4 } { 1 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { 4 } { 1 }$
$ $ 0 has no sign $ $
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { 4 } { 1 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { 4 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ 4 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-If the sum of two consecutive 
numbers is $45$ and one number is $X$ 
.This statement in the form of 
equation $1s:$ 
$\left(1$ Point) $\right)$ 
$○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ 
$○sx+1ef\left(x+2$ $r1gnt\right)=145s$ 
$sx+1x=45s$
7th-9th grade
Algebra
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ 
$s1S$ 
$S-1S$ 
$s2S$ 
$s50s$
7th-9th grade
Other
search-thumbnail-Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ 
Find f(7)fAleft(7\right) 
O a 
O b -8 
6. 
$5$
7th-9th grade
Algebra
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