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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
$y = 5 x ^ { 2 } - 12$
$y = 0$
$x$-intercept
$\left ( \dfrac { 2 \sqrt{ 15 } } { 5 } , 0 \right )$, $\left ( - \dfrac { 2 \sqrt{ 15 } } { 5 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 12 \right )$
Minimum
$\left ( 0 , - 12 \right )$
Standard form
$y = 5 x ^ { 2 } - 12$
$5x ^{ 2 } -12 = 0$
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = - \dfrac { 2 \sqrt{ 15 } } { 5 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } } = \color{#FF6800}{ 0 }$
$x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$x ^ { 2 } = \color{#FF6800}{ \dfrac { 12 } { 5 } }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 12 } { 5 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 12 } { 5 } } }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 12 } { 5 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \end{array}$
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = - \dfrac { 2 \sqrt{ 15 } } { 5 } \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$
$ $ 0 has no sign $ $
$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$
$ $ The power of 0 is 0 $ $
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 0 - 4 \times 5 \times \left ( - 12 \right ) } } { 2 \times 5 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 240 } } { 2 \times 5 } }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 240 } } } { 2 \times 5 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 0 \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 15 } } } { 2 \times 5 }$
$x = \dfrac { 0 \pm 4 \sqrt{ 15 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } }$
$ $ Multiply $ 2 $ and $ 5$
$x = \dfrac { 0 \pm 4 \sqrt{ 15 } } { \color{#FF6800}{ 10 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm 4 \sqrt{ 15 } } { 10 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 4 \sqrt{ 15 } } { 10 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 0 } + 4 \sqrt{ 15 } } { 10 } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = \dfrac { 4 \sqrt{ 15 } } { 10 } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 4 \sqrt{ 15 } } { 10 } } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \\ x = \dfrac { 0 - 4 \sqrt{ 15 } } { 10 } \end{array}$
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \dfrac { \color{#FF6800}{ 0 } - 4 \sqrt{ 15 } } { 10 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \dfrac { - 4 \sqrt{ 15 } } { 10 } \end{array}$
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ \dfrac { - 4 \sqrt{ 15 } } { 10 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 15 } } { 5 } } \end{array}$
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 15 } } { 5 } } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = \dfrac { 2 \sqrt{ 15 } } { 5 } \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 15 } } { 5 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times \left ( - 12 \right )$
$ $ The power of 0 is 0 $ $
$D = \color{#FF6800}{ 0 } - 4 \times 5 \times \left ( - 12 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 5 \times \left ( - 12 \right )$
$ $ 0 does not change when you add or subtract $ $
$D = - 4 \times 5 \times \left ( - 12 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$
$ $ Multiply the numbers $ $
$D = \color{#FF6800}{ 240 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 240 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 0 , \alpha \beta = - \dfrac { 12 } { 5 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 5 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 12 } { 5 } }$
$\alpha + \beta = - \color{#FF6800}{ \dfrac { 0 } { 5 } } , \alpha \beta = \dfrac { - 12 } { 5 }$
$ $ If the numerator is 0, it is equal to 0 $ $
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 5 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 5 }$
$ $ 0 has no sign $ $
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 5 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 12 } } { 5 }$
$ $ Move the minus sign to the front of the fraction $ $
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-
PRACTICE TASK $\times 3$ teaching purpose 
List all the possible zeros of each Notfor sale 
polynomial. 
$1$ $γ3+9y^{2}-44=0$ 
$2$ $x^{4}-3x^{2}+2=0$ 
$3$ $2x^{2}-5x-3=0$ 
$4$ $x^{3}-2x^{2}-11x+12=0$ 
$5$ $x^{4}-2x^{2}-16x-15=0$
10th-13th grade
Trigonometry
search-thumbnail-c. Factoring 
$1$ $n^{2}-5n=0$ 
$2$ $d^{2}+2d-15=0$ 
$3$ $2c^{2}-8c+8=0$ 
$4$ $4g^{2}-12g+9=0$ 
$5$ $x^{2}$ $-3x=4$
7th-9th grade
Other
search-thumbnail-$cti$ $y\right)$ It's in My Roots! 
Solve the following quadratic equations. 
$1$ $x^{2}-16=0$ $6.$ $\left(x-2\right)^{2}-64=0$ 
$2.$ $z^{2}+256=0$ $7.$ $\left(x+5\right)^{2}-81=0$ 
$3.$ $x^{2}-121=0$ $8.$ $\left(x-1\right)^{2}-24=0$ 

$4.$ $3x^{2}-27=0$ 
$9.$ $\left(x+2\right)^{2}-32=0$ 
$5.$ $5x^{2}-125=0$ 
$10.\left(x-2\right)^{2}-12=0$
7th-9th grade
Other
search-thumbnail-List all the possible zeros of each por 
$1$ $y^{3}+9y^{2}-44=0$ 
$2$ $x^{4}-3x^{2}+2=0$ 
$3$ $2x^{2}-5x-3=0$ 
$4$ $x^{3}-2x^{2}-11x+12=0$ 
$5$ $x^{4}-2x^{2}-16x-15=0$
10th-13th grade
Other
search-thumbnail-Short Answer 
1. Use Newton's Method to find the desired approximation for 
$2x^{2}-12=05x=2x_{c}=2$ 
$f$ $\left(x\right)=x^{2}-12$ $x_{1}=2$ 
$f\left(x\right)=2\times $ $k_{2}=$ 
$t3$ 
$-$ $\dfrac {x^{2}-12} {1<}$ 
$-$ $13$ $4=3.HC$ 
$b.$ $x^{3}4-5x-5=θx_{1}=2x_{3}=2$ 
$x^{3}-5x-5=c5\times ,=z:x^{22}$ 
$\times 3=2.68$ 
$c.4$ $si_{nx}=05x_{1}=35x_{3}=2\left(x$ is in radians) 
$\times 3=3.14$
10th-13th grade
Calculus
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