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Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = 3 x ^ { 2 } + 5 x + 1$
$y = 0$
$x$Intercept
$\left ( - \dfrac { 5 } { 6 } + \dfrac { \sqrt{ 13 } } { 6 } , 0 \right )$, $\left ( - \dfrac { 5 } { 6 } - \dfrac { \sqrt{ 13 } } { 6 } , 0 \right )$
$y$Intercept
$\left ( 0 , 1 \right )$
Minimum
$\left ( - \dfrac { 5 } { 6 } , - \dfrac { 13 } { 12 } \right )$
Standard form
$y = 3 \left ( x + \dfrac { 5 } { 6 } \right ) ^ { 2 } - \dfrac { 13 } { 12 }$
$3x ^{ 2 } +5x+1 = 0$
$\begin{array} {l} x = \dfrac { - 5 + \sqrt{ 13 } } { 6 } \\ x = \dfrac { - 5 - \sqrt{ 13 } } { 6 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
 Divide both sides by the coefficient of the leading highest term 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 5 } { 6 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x + \dfrac { 5 } { 6 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 5 } { 6 } \right ) ^ { 2 } = - \dfrac { 1 } { 3 } + \left ( \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } }$
 When raising a fraction to the power, raise the numerator and denominator each to the power 
$\left ( x + \dfrac { 5 } { 6 } \right ) ^ { 2 } = - \dfrac { 1 } { 3 } + \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 ^ { 2 } } { 6 ^ { 2 } } }$
 Organize the expression 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 13 } { 36 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 6 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 13 } { 36 } }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 6 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 13 } { 36 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 6 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 13 } { 36 } } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 13 } } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 13 } } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 13 } } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 13 } } { 6 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 13 } } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 13 } } { 6 } } \end{array}$
 Organize the expression 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 + \sqrt{ 13 } } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 - \sqrt{ 13 } } { 6 } } \end{array}$
$\begin{array} {l} x = \dfrac { - 5 + \sqrt{ 13 } } { 6 } \\ x = \dfrac { - 5 - \sqrt{ 13 } } { 6 } \end{array}$
$x = \dfrac { - 5 \pm \sqrt{ \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 1 } } { 2 \times 3 }$
 Calculate power 
$x = \dfrac { - 5 \pm \sqrt{ \color{#FF6800}{ 25 } - 4 \times 3 \times 1 } } { 2 \times 3 }$
$x = \dfrac { - 5 \pm \sqrt{ 25 - 4 \times 3 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } } { 2 \times 3 }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { - 5 \pm \sqrt{ 25 - 4 \times 3 } } { 2 \times 3 }$
$x = \dfrac { - 5 \pm \sqrt{ 25 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } } } { 2 \times 3 }$
 Multiply $- 4$ and $3$
$x = \dfrac { - 5 \pm \sqrt{ 25 \color{#FF6800}{ - } \color{#FF6800}{ 12 } } } { 2 \times 3 }$
$x = \dfrac { - 5 \pm \sqrt{ \color{#FF6800}{ 25 } \color{#FF6800}{ - } \color{#FF6800}{ 12 } } } { 2 \times 3 }$
 Subtract $12$ from $25$
$x = \dfrac { - 5 \pm \sqrt{ \color{#FF6800}{ 13 } } } { 2 \times 3 }$
$x = \dfrac { - 5 \pm \sqrt{ 13 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } }$
 Multiply $2$ and $3$
$x = \dfrac { - 5 \pm \sqrt{ 13 } } { \color{#FF6800}{ 6 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 \pm \sqrt{ 13 } } { 6 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 + \sqrt{ 13 } } { 6 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 5 - \sqrt{ 13 } } { 6 } } \end{array}$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
$D = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } - 4 \times 3 \times 1$
 Calculate power 
$D = \color{#FF6800}{ 25 } - 4 \times 3 \times 1$
$D = 25 - 4 \times 3 \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
 Multiplying any number by 1 does not change the value 
$D = 25 - 4 \times 3$
$D = 25 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 }$
 Multiply $- 4$ and $3$
$D = 25 \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$D = \color{#FF6800}{ 25 } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
 Subtract $12$ from $25$
$D = \color{#FF6800}{ 13 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 13 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = - \dfrac { 5 } { 3 } , \alpha \beta = \dfrac { 1 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
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