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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
$y = x ^ { 2 } - 36$
$y = 0$
$x$-intercept
$\left ( - 6 , 0 \right )$, $\left ( 6 , 0 \right )$
$y$-intercept
$\left ( 0 , - 36 \right )$
Minimum
$\left ( 0 , - 36 \right )$
Standard form
$y = x ^ { 2 } - 36$
$x ^{ 2 } -36 = 0$
$\begin{array} {l} x = 6 \\ x = - 6 \end{array}$
Solve quadratic equations using the square root
$x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 36 } = 0$
$ $ Move the constant to the right side and change the sign $ $
$x ^ { 2 } = \color{#FF6800}{ 36 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 36 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 36 } }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 36 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 6 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 6 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \end{array}$
$\begin{array} {l} x = 6 \\ x = - 6 \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 36 \right ) } } { 2 \times 1 }$
$ $ 0 has no sign $ $
$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 36 \right ) } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 36 \right ) } } { 2 \times 1 }$
$ $ The power of 0 is 0 $ $
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 36 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 0 - 4 \times 1 \times \left ( - 36 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 144 } } { 2 \times 1 } }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 144 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 0 \pm \color{#FF6800}{ 12 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm 12 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 0 \pm 12 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm 12 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 12 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 12 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 0 } + 12 } { 2 } \\ x = \dfrac { 0 - 12 } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = \dfrac { 12 } { 2 } \\ x = \dfrac { 0 - 12 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 12 } { 2 } } \\ x = \dfrac { 0 - 12 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 } { 1 } } \\ x = \dfrac { 0 - 12 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 } { 1 } } \\ x = \dfrac { 0 - 12 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 6 } \\ x = \dfrac { 0 - 12 } { 2 } \end{array}$
$\begin{array} {l} x = 6 \\ x = \dfrac { \color{#FF6800}{ 0 } - 12 } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = 6 \\ x = \dfrac { - 12 } { 2 } \end{array}$
$\begin{array} {l} x = 6 \\ x = \color{#FF6800}{ \dfrac { - 12 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 6 \\ x = \color{#FF6800}{ \dfrac { - 6 } { 1 } } \end{array}$
$\begin{array} {l} x = 6 \\ x = \dfrac { - 6 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 6 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 36 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 36 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 36 \right )$
$ $ The power of 0 is 0 $ $
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 36 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 36 \right )$
$ $ 0 does not change when you add or subtract $ $
$D = - 4 \times 1 \times \left ( - 36 \right )$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 36 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = - 4 \times \left ( - 36 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 36 } \right )$
$ $ Multiply $ - 4 $ and $ - 36$
$D = \color{#FF6800}{ 144 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 144 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 0 , \alpha \beta = - 36$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 36 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 36 } { 1 } }$
$\alpha + \beta = - \dfrac { 0 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 36 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 36 } { 1 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 36 } { 1 }$
$ $ 0 has no sign $ $
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 36 } { 1 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { - 36 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 36 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-The locus of the point which is at a distance 
6units from $y-a\times 1s$ is 
$x^{2}$ $-25=0$ 
$x^{2}-36=0$ 
$y+36=0$ 
$y^{2}-36=0$
10th-13th grade
Algebra
search-thumbnail-$1$ $x^{3}-2x^{2}-3x+6=0$ $x=1$ 
$2$ $x^{4}-5x^{2}$ $-36=0$ $x=3$ 
3. $x^{3}+3x^{2}-x-3=0$ $x=$ = $-3$ 
$4.$ $x^{3}+3x^{2}$ $-14x-20=0$ $x=1$ 
$5$ $2x^{4}$ $-17x^{3}+13x^{2}+53x+21$ $x=1$
10th-13th grade
Other
search-thumbnail-$1°$ $x^{4}+5x^{2}-36=0$ 
$2°$ $x^{4}-25x^{2}+144=0$ 
$3$ $\left(x^{2}+x\right)^{2}-\left(x^{2}+x\right)-2=0$ 
$2$ 
$4a$ $\left(\dfrac {x-2} {x+2}\right)^{2}-4\left(\dfrac {x-2} {x+2}\right)+3=0,x≠2$
10th-13th grade
Other
search-thumbnail-Auer $4-8$ $\dfrac {x} {18}$ $\dfrac {x3} {18x}|=\dfrac {6} {18}$ $\bar{6} $ 
$x^{2}-36=36-36$ 
$x^{2}-36=0$ 
$x^{2}=36$ 
$x=±6$ $\left(8\right)$ $3$
10th-13th grade
Other
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