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Solve the quadratic equation
Answer
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Number of solution
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Relationship between roots and coefficients
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Graph
$y = 0.6 x ^ { 2 } - \dfrac { x ^ { 2 } - x } { 5 }$
$y = 1$
$x$Intercept
$\left ( 0 , 0 \right )$, $\left ( - \dfrac { 1 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
$0.6x ^{ 2 } - \dfrac{ x ^{ 2 } -x }{ 5 } = 1$
$\begin{array} {l} x = \dfrac { - 1 + \sqrt{ 41 } } { 4 } \\ x = \dfrac { - 1 - \sqrt{ 41 } } { 4 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 0.6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - x } { 5 } } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = 1$
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } = \color{#FF6800}{ 5 }$
$2 x ^ { 2 } + x = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 1 } { 4 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 1 } { 4 } \right ) ^ { 2 } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 1 } { 4 } \right ) ^ { 2 } = \dfrac { 5 } { 2 } + \left ( \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 1 } { 4 } \right ) ^ { 2 } = \dfrac { 5 } { 2 } + \dfrac { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 ^ { 2 } } { 4 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 41 } { 16 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 41 } { 16 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 41 } { 16 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 41 } { 16 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 41 } } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 41 } } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 41 } } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 41 } } { 4 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 41 } } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 41 } } { 4 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 + \sqrt{ 41 } } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 - \sqrt{ 41 } } { 4 } } \end{array}$
$\begin{array} {l} x = \dfrac { - 1 + \sqrt{ 41 } } { 4 } \\ x = \dfrac { - 1 - \sqrt{ 41 } } { 4 } \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ 0.6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - x } { 5 } } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = 1$
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } = \color{#FF6800}{ 5 }$
$2 x ^ { 2 } + x = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$x = \dfrac { - 1 \pm \sqrt{ \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 5 \right ) } } { 2 \times 2 }$
$ $ Calculate power $ $
$x = \dfrac { - 1 \pm \sqrt{ \color{#FF6800}{ 1 } - 4 \times 2 \times \left ( - 5 \right ) } } { 2 \times 2 }$
$x = \dfrac { - 1 \pm \sqrt{ 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) } } { 2 \times 2 }$
$ $ Multiply the numbers $ $
$x = \dfrac { - 1 \pm \sqrt{ 1 + \color{#FF6800}{ 40 } } } { 2 \times 2 }$
$x = \dfrac { - 1 \pm \sqrt{ \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 40 } } } { 2 \times 2 }$
$ $ Add $ 1 $ and $ 40$
$x = \dfrac { - 1 \pm \sqrt{ \color{#FF6800}{ 41 } } } { 2 \times 2 }$
$x = \dfrac { - 1 \pm \sqrt{ 41 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$ $ Multiply $ 2 $ and $ 2$
$x = \dfrac { - 1 \pm \sqrt{ 41 } } { \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 \pm \sqrt{ 41 } } { 4 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 + \sqrt{ 41 } } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 1 - \sqrt{ 41 } } { 4 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 0.6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - x } { 5 } } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = 1$
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } = \color{#FF6800}{ 5 }$
$2 x ^ { 2 } + x = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 5 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1 } - 4 \times 2 \times \left ( - 5 \right )$
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$ $ Multiply the numbers $ $
$D = 1 + \color{#FF6800}{ 40 }$
$D = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 40 }$
$ $ Add $ 1 $ and $ 40$
$D = \color{#FF6800}{ 41 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 41 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - \dfrac { 1 } { 2 } , \alpha \beta = - \dfrac { 5 } { 2 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 0.6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - x } { 5 } } = 1$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = 1$
$\color{#FF6800}{ \dfrac { 2 x ^ { 2 } + x } { 5 } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } = \color{#FF6800}{ 5 }$
$2 x ^ { 2 } + x = \color{#FF6800}{ 5 }$
$ $ Move the expression to the left side and change the symbol $ $
$2 x ^ { 2 } + x \color{#FF6800}{ - } \color{#FF6800}{ 5 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 5 } { 2 } }$
$\alpha + \beta = - \dfrac { 1 } { 2 } , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { 2 }$
$ $ Move the minus sign to the front of the fraction $ $
$\alpha + \beta = - \dfrac { 1 } { 2 } , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
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