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Solve the quadratic equation
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Number of solution
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Relationship between roots and coefficients
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Graph
$y = \dfrac { \left ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } - 2 x \left ( x - 1 \right )$
$y = 0$
$y$Intercept
$\left ( 0 , - 5 \right )$
$\begin{array} {l} x = \dfrac { 7 + \sqrt{ 11 } i } { 2 } \\ x = \dfrac { 7 - \sqrt{ 11 } i } { 2 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ \dfrac { \left ( \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$ $ Arrange the fraction expression $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$ $ Expand the expression $ $
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 0$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = - 15 + \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = - 15 + \dfrac { \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 11 } } { \color{#FF6800}{ 4 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 11 } } { \color{#FF6800}{ 4 } } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 11 } } { \color{#FF6800}{ 4 } } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 11 } } { \color{#FF6800}{ 4 } } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} x = \dfrac { 7 + \sqrt{ 11 } i } { 2 } \\ x = \dfrac { 7 - \sqrt{ 11 } i } { 2 } \end{array}$
Calculate using the quodratic formula$($Imaginary root solution$)
$\color{#FF6800}{ \dfrac { \left ( \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$ $ Arrange the fraction expression $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$ $ Expand the expression $ $
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 0$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 7 \right ) \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 7 \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$x = \dfrac { 7 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 7 \pm \sqrt{ 7 ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \pm \sqrt{ \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 11 } } } { \color{#FF6800}{ 2 } } }$
$x = \dfrac { 7 \pm \sqrt{ \color{#FF6800}{ - } 11 } } { 2 }$
$ $ Subtracting (-) from the square root gives i $ $
$x = \dfrac { 7 \pm \sqrt{ 11 } \color{#FF6800}{ i } } { 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \pm \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 11 } } \color{#FF6800}{ i } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Do not have the solution $ $
Calculate using the quadratic formula
$\color{#FF6800}{ \dfrac { \left ( \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$ $ Arrange the fraction expression $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$ $ Expand the expression $ $
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 0$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 7 \right ) \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 7 \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$x = \dfrac { 7 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 7 \pm \sqrt{ 7 ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \pm \sqrt{ \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 11 } } } { \color{#FF6800}{ 2 } } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 11 } } } { \color{#FF6800}{ 2 } } }$
$ $ The square root of a negative number does not exist within the set of real numbers $ $
$ $ Do not have the solution $ $
$ $ Do not have the real root $ $
Find the number of solutions
$\color{#FF6800}{ \dfrac { \left ( \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$ $ Arrange the fraction expression $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$ $ Expand the expression $ $
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 0$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 15$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 7 ^ { 2 } - 4 \times 1 \times 15$
$D = \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 15$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 49 } - 4 \times 1 \times 15$
$D = 49 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 15$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 49 - 4 \times 15$
$D = 49 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 }$
$ $ Multiply $ - 4 $ and $ 15$
$D = 49 \color{#FF6800}{ - } \color{#FF6800}{ 60 }$
$D = \color{#FF6800}{ 49 } \color{#FF6800}{ - } \color{#FF6800}{ 60 }$
$ $ Subtract $ 60 $ from $ 49$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 11 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ - } \color{#FF6800}{ 11 }$
$ $ Since $ D<0 $ , there is no real root of the following quadratic equation $ $
$ $ Do not have the real root $ $
$\alpha + \beta = 7 , \alpha \beta = 15$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ \dfrac { \left ( \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 11 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$ $ Arrange the fraction expression $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } - 2 x \left ( x - 1 \right ) = 0$
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$ $ Expand the expression $ $
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ 0 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 0$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 7 } } { \color{#FF6800}{ 1 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 15 } } { \color{#FF6800}{ 1 } } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 7 } } { \color{#FF6800}{ 1 } } } , \alpha \beta = \dfrac { 15 } { 1 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 7 } } { \color{#FF6800}{ 1 } } } , \alpha \beta = \dfrac { 15 } { 1 }$
$\alpha + \beta = \dfrac { 7 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 15 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = \color{#FF6800}{ 7 } , \alpha \beta = \dfrac { 15 } { 1 }$
$\alpha + \beta = 7 , \alpha \beta = \dfrac { 15 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 7 , \alpha \beta = \color{#FF6800}{ 15 }$
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