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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
$y = x ^ { 2 } - 25$
$y = 0$
$x$Intercept
$\left ( - 5 , 0 \right )$, $\left ( 5 , 0 \right )$
$y$Intercept
$\left ( 0 , - 25 \right )$
Minimum
$\left ( 0 , - 25 \right )$
Standard form
$y = x ^ { 2 } - 25$
$x ^{ 2 } -25 = 0$
$\begin{array} {l} x = 5 \\ x = - 5 \end{array}$
Solve quadratic equations using the square root
$x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
$ $ Move the constant to the right side and change the sign $ $
$x ^ { 2 } = \color{#FF6800}{ 25 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 25 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 25 } }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 25 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$\begin{array} {l} x = 5 \\ x = - 5 \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 25 \right ) } } { 2 \times 1 }$
$ $ 0 has no sign $ $
$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 25 \right ) } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 25 \right ) } } { 2 \times 1 }$
$ $ The power of 0 is 0 $ $
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 25 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 0 - 4 \times 1 \times \left ( - 25 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 100 } } { 2 \times 1 } }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 100 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 0 \pm \color{#FF6800}{ 10 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm 10 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 0 \pm 10 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm 10 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 10 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 10 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 0 } + 10 } { 2 } \\ x = \dfrac { 0 - 10 } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = \dfrac { 10 } { 2 } \\ x = \dfrac { 0 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 10 } { 2 } } \\ x = \dfrac { 0 - 10 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 0 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 0 - 10 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 5 } \\ x = \dfrac { 0 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 0 } - 10 } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = 5 \\ x = \dfrac { - 10 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { - 10 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { - 5 } { 1 } } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { - 5 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 25 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 25 \right )$
$ $ The power of 0 is 0 $ $
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 25 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 25 \right )$
$ $ 0 does not change when you add or subtract $ $
$D = - 4 \times 1 \times \left ( - 25 \right )$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 25 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = - 4 \times \left ( - 25 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 25 } \right )$
$ $ Multiply $ - 4 $ and $ - 25$
$D = \color{#FF6800}{ 100 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 0 , \alpha \beta = - 25$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 25 } { 1 } }$
$\alpha + \beta = - \dfrac { 0 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 25 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 25 } { 1 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 25 } { 1 }$
$ $ 0 has no sign $ $
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 25 } { 1 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { - 25 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 25 }$
$ $ 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-$x^{2}$ $-6x-27=0$ 
$x^{2}-25=0$ 
$x^{2+10x+25}=0$ 
$2x^{2}-5x+3=0$
7th-9th grade
Other
search-thumbnail-$1y$ Complete the table. 
Roots $x1$ $+\times 2$ $1$ $xy$ 
Equation $7$ $1\right)$ $a$ $1$ $2$ 
$x^{2+5x+4}=0$ 
$x^{2-6x-27}=0$ 
$x^{2}-25=0$ 
$x^{2+10x+25}=0$ 
$2x^{2}-5x+3=0$
7th-9th grade
Other
search-thumbnail-Equalion 
$X^{2}$ $-6x-27=0$ 
$x^{2}-25=0$ 
$x^{2+10x+25}=0$ 
$2x^{2}-5x+3=0$
7th-9th grade
Other
search-thumbnail-B. Complete the table. 
Roots 
Equation $a$ $b$ $C$ $x1$ $\times 2$ $\times 1$ $+\times 2$ $\times 1$ $.\times 2$ 
$x^{2+5x+4}=0$ 
$x^{2-6x-27}=0$ 
$x^{2}-25=0$ 
$x^{2+10x+25}=0$ 
$2x^{2}-5x+3=0$
7th-9th grade
Other
search-thumbnail-Now, let's try solving these using square roots: 
$x^{2}-25=0$ $x^{2}-81=0$
10th-13th grade
Other
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