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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
$y = x ^ { 2 } - 6 x - 9$
$y = 0$
$x$-intercept
$\left ( 3 - 3 \sqrt{ 2 } , 0 \right )$, $\left ( 3 + 3 \sqrt{ 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 9 \right )$
Minimum
$\left ( 3 , - 18 \right )$
Standard form
$y = \left ( x - 3 \right ) ^ { 2 } - 18$
$x ^{ 2 } -6x-9 = 0$
$\begin{array} {l} x = 3 + 3 \sqrt{ 2 } \\ x = 3 - 3 \sqrt{ 2 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - 3 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - 3 \right ) ^ { 2 } = \color{#FF6800}{ 9 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x - 3 \right ) ^ { 2 } = 9 + \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } }$
$ $ Calculate power $ $
$\left ( x - 3 \right ) ^ { 2 } = 9 + \color{#FF6800}{ 9 }$
$\left ( x - 3 \right ) ^ { 2 } = \color{#FF6800}{ 9 } \color{#FF6800}{ + } \color{#FF6800}{ 9 }$
$ $ Add $ 9 $ and $ 9$
$\left ( x - 3 \right ) ^ { 2 } = \color{#FF6800}{ 18 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 18 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \pm \sqrt{ \color{#FF6800}{ 18 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \pm \sqrt{ \color{#FF6800}{ 18 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \end{array}$
$\begin{array} {l} x = 3 + 3 \sqrt{ 2 } \\ x = 3 - 3 \sqrt{ 2 } \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 6 \right ) \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 6 \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 }$
$x = \dfrac { 6 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm \sqrt{ 72 } } { 2 \times 1 } }$
$x = \dfrac { 6 \pm \sqrt{ \color{#FF6800}{ 72 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 6 \pm \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 2 } } } { 2 \times 1 }$
$x = \dfrac { 6 \pm 6 \sqrt{ 2 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 6 \pm 6 \sqrt{ 2 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm 6 \sqrt{ 2 } } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 + 6 \sqrt{ 2 } } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 - 6 \sqrt{ 2 } } { 2 } } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 + 6 \sqrt{ 2 } } { 2 } } \\ x = \dfrac { 6 - 6 \sqrt{ 2 } } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 + 3 \sqrt{ 2 } } { 1 } } \\ x = \dfrac { 6 - 6 \sqrt{ 2 } } { 2 } \end{array}$
$\begin{array} {l} x = \dfrac { 3 + 3 \sqrt{ 2 } } { \color{#FF6800}{ 1 } } \\ x = \dfrac { 6 - 6 \sqrt{ 2 } } { 2 } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \\ x = \dfrac { 6 - 6 \sqrt{ 2 } } { 2 } \end{array}$
$\begin{array} {l} x = 3 + 3 \sqrt{ 2 } \\ x = \color{#FF6800}{ \dfrac { 6 - 6 \sqrt{ 2 } } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 3 + 3 \sqrt{ 2 } \\ x = \color{#FF6800}{ \dfrac { 3 - 3 \sqrt{ 2 } } { 1 } } \end{array}$
$\begin{array} {l} x = 3 + 3 \sqrt{ 2 } \\ x = \dfrac { 3 - 3 \sqrt{ 2 } } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 3 + 3 \sqrt{ 2 } \\ x = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 9 \right )$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 6 ^ { 2 } - 4 \times 1 \times \left ( - 9 \right )$
$D = \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 9 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 36 } - 4 \times 1 \times \left ( - 9 \right )$
$D = 36 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 9 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 36 - 4 \times \left ( - 9 \right )$
$D = 36 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$ $ Multiply $ - 4 $ and $ - 9$
$D = 36 + \color{#FF6800}{ 36 }$
$D = \color{#FF6800}{ 36 } \color{#FF6800}{ + } \color{#FF6800}{ 36 }$
$ $ Add $ 36 $ and $ 36$
$D = \color{#FF6800}{ 72 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 72 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 6 , \alpha \beta = - 9$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 6 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 9 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 6 } { 1 } } , \alpha \beta = \dfrac { - 9 } { 1 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 6 } { 1 } } , \alpha \beta = \dfrac { - 9 } { 1 }$
$\alpha + \beta = \dfrac { 6 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 9 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = \color{#FF6800}{ 6 } , \alpha \beta = \dfrac { - 9 } { 1 }$
$\alpha + \beta = 6 , \alpha \beta = \dfrac { - 9 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 6 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 9 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-
Find the radical centre of the following circles. 
$\left(1\right)$ $x^{2}+y^{2}-4x-6y+5=0$ $x^{2}+y^{2}-2x-4y-1=0,$ $x^{2}+y^{2}-6x-2y=0$ 
$\left(ii\right)$ $x^{2}+y^{2}+4x-7=0$ $2x^{2}+2y^{2}+3x+5y-9=0$ $x^{2}+y^{2}+y=0$
10th-13th grade
Other
search-thumbnail-$4$ $x^{2}+y^{2}-6x-9y.+13=0$ 
$x^{2}+y^{2}-2x-16y$ $=$ $c$
10th-13th grade
Other
search-thumbnail-$x^{2}-6x-9=0$
10th-13th grade
Other
search-thumbnail-Q. $48$ of $60$ Mark for review 
Circle(s) 
$\left(s\right)$ touching the $x-a\times 1S$ at a distance 
of $3$ units from the origin and having an 
intercept of length $2\sqrt{7} $ on the $y-$ 
$y-a\times 1s$ is 
$\left($ (are) 
$2\right)$ 
$x^{2}+y^{2}-6x+8y+9=0$ 
$x^{2}+y^{2}-6x+7y+9=0$ 
$x^{2}+y^{2}-6x-8y+9=0$ 
$x^{2}+y^{2}-6x-7y+9=0$
10th-13th grade
Algebra
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