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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
$y = x ^ { 2 } + 2 x - 15$
$y = 0$
$x$Intercept
$\left ( - 5 , 0 \right )$, $\left ( 3 , 0 \right )$
$y$Intercept
$\left ( 0 , - 15 \right )$
Minimum
$\left ( - 1 , - 16 \right )$
Standard form
$y = \left ( x + 1 \right ) ^ { 2 } - 16$
$x ^{ 2 } +2x-15 = 0$
$\begin{array} {l} x = 3 \\ x = - 5 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$\begin{array} {l} x = 3 \\ x = - 5 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + 1 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x + 1 \right ) ^ { 2 } = 15 + \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
$ $ Calculate power $ $
$\left ( x + 1 \right ) ^ { 2 } = 15 + \color{#FF6800}{ 1 }$
$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Add $ 15 $ and $ 1$
$\left ( x + 1 \right ) ^ { 2 } = \color{#FF6800}{ 16 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 16 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ 16 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \pm \sqrt{ \color{#FF6800}{ 16 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \\ x = - 1 - 4 \end{array}$
$ $ Add $ - 1 $ and $ 4$
$\begin{array} {l} x = \color{#FF6800}{ 3 } \\ x = - 1 - 4 \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$\begin{array} {l} x = 3 \\ x = - 5 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ 2 ^ { 2 } - 4 \times 1 \times \left ( - 15 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm \sqrt{ 64 } } { 2 \times 1 } }$
$x = \dfrac { - 2 \pm \sqrt{ \color{#FF6800}{ 64 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { - 2 \pm \color{#FF6800}{ 8 } } { 2 \times 1 }$
$x = \dfrac { - 2 \pm 8 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { - 2 \pm 8 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \pm 8 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 + 8 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 - 8 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } } { 2 } \\ x = \dfrac { - 2 - 8 } { 2 } \end{array}$
$ $ Add $ - 2 $ and $ 8$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 6 } } { 2 } \\ x = \dfrac { - 2 - 8 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 } { 2 } } \\ x = \dfrac { - 2 - 8 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \\ x = \dfrac { - 2 - 8 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \\ x = \dfrac { - 2 - 8 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 3 } \\ x = \dfrac { - 2 - 8 } { 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 8 } } { 2 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = 3 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 2 } \end{array}$
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { - 10 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ \dfrac { - 5 } { 1 } } \end{array}$
$\begin{array} {l} x = 3 \\ x = \dfrac { - 5 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 3 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right )$
$D = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 15 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 4 } - 4 \times 1 \times \left ( - 15 \right )$
$D = 4 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 15 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 4 - 4 \times \left ( - 15 \right )$
$D = 4 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right )$
$ $ Multiply $ - 4 $ and $ - 15$
$D = 4 + \color{#FF6800}{ 60 }$
$D = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 60 }$
$ $ Add $ 4 $ and $ 60$
$D = \color{#FF6800}{ 64 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 64 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - 2 , \alpha \beta = - 15$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 15 } { 1 } }$
$\alpha + \beta = - \dfrac { 2 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 15 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 2 } , \alpha \beta = \dfrac { - 15 } { 1 }$
$\alpha + \beta = - 2 , \alpha \beta = \dfrac { - 15 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - 2 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 15 }$
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