# Calculator search results

Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = x ^ { 2 } - 8 x + 3$
$y = 0$
$x$-intercept
$\left ( 4 - \sqrt{ 13 } , 0 \right )$, $\left ( \sqrt{ 13 } + 4 , 0 \right )$
$y$-intercept
$\left ( 0 , 3 \right )$
Minimum
$\left ( 4 , - 13 \right )$
Standard form
$y = \left ( x - 4 \right ) ^ { 2 } - 13$
$x ^{ 2 } -8x+3 = 0$
$\begin{array} {l} x = 4 + \sqrt{ 13 } \\ x = 4 - \sqrt{ 13 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - 4 \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x - 4 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x - 4 \right ) ^ { 2 } = - 3 + \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } }$
 Calculate power 
$\left ( x - 4 \right ) ^ { 2 } = - 3 + \color{#FF6800}{ 16 }$
$\left ( x - 4 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
 Add $- 3$ and $16$
$\left ( x - 4 \right ) ^ { 2 } = \color{#FF6800}{ 13 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 13 }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 13 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 13 } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 13 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 13 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 13 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 13 } } \end{array}$
$\begin{array} {l} x = 4 + \sqrt{ 13 } \\ x = 4 - \sqrt{ 13 } \end{array}$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 8 \right ) \pm \sqrt{ \left ( - 8 \right ) ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 }$
 Simplify Minus 
$x = \dfrac { 8 \pm \sqrt{ \left ( - 8 \right ) ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 }$
$x = \dfrac { 8 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 8 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 3 } } { 2 \times 1 }$
 Remove negative signs because negative numbers raised to even powers are positive 
$x = \dfrac { 8 \pm \sqrt{ 8 ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 \pm \sqrt{ 8 ^ { 2 } - 4 \times 1 \times 3 } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 \pm \sqrt{ 52 } } { 2 \times 1 } }$
$x = \dfrac { 8 \pm \sqrt{ \color{#FF6800}{ 52 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$x = \dfrac { 8 \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 13 } } } { 2 \times 1 }$
$x = \dfrac { 8 \pm 2 \sqrt{ 13 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { 8 \pm 2 \sqrt{ 13 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 \pm 2 \sqrt{ 13 } } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 + 2 \sqrt{ 13 } } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 - 2 \sqrt{ 13 } } { 2 } } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 8 + 2 \sqrt{ 13 } } { 2 } } \\ x = \dfrac { 8 - 2 \sqrt{ 13 } } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 4 + \sqrt{ 13 } } { 1 } } \\ x = \dfrac { 8 - 2 \sqrt{ 13 } } { 2 } \end{array}$
$\begin{array} {l} x = \dfrac { 4 + \sqrt{ 13 } } { \color{#FF6800}{ 1 } } \\ x = \dfrac { 8 - 2 \sqrt{ 13 } } { 2 } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} x = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 13 } } \\ x = \dfrac { 8 - 2 \sqrt{ 13 } } { 2 } \end{array}$
$\begin{array} {l} x = 4 + \sqrt{ 13 } \\ x = \color{#FF6800}{ \dfrac { 8 - 2 \sqrt{ 13 } } { 2 } } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = 4 + \sqrt{ 13 } \\ x = \color{#FF6800}{ \dfrac { 4 - \sqrt{ 13 } } { 1 } } \end{array}$
$\begin{array} {l} x = 4 + \sqrt{ 13 } \\ x = \dfrac { 4 - \sqrt{ 13 } } { \color{#FF6800}{ 1 } } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} x = 4 + \sqrt{ 13 } \\ x = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 13 } } \end{array}$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 8 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 8 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 3$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 8 ^ { 2 } - 4 \times 1 \times 3$
$D = \color{#FF6800}{ 8 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 3$
 Calculate power 
$D = \color{#FF6800}{ 64 } - 4 \times 1 \times 3$
$D = 64 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 3$
 Multiplying any number by 1 does not change the value 
$D = 64 - 4 \times 3$
$D = 64 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 }$
 Multiply $- 4$ and $3$
$D = 64 \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$D = \color{#FF6800}{ 64 } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
 Subtract $12$ from $64$
$D = \color{#FF6800}{ 52 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 52 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = 8 , \alpha \beta = 3$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 8 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 3 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 8 } { 1 } } , \alpha \beta = \dfrac { 3 } { 1 }$
 Solve the sign of a fraction with a negative sign 
$\alpha + \beta = \color{#FF6800}{ \dfrac { 8 } { 1 } } , \alpha \beta = \dfrac { 3 } { 1 }$
$\alpha + \beta = \dfrac { 8 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 3 } { 1 }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = \color{#FF6800}{ 8 } , \alpha \beta = \dfrac { 3 } { 1 }$
$\alpha + \beta = 8 , \alpha \beta = \dfrac { 3 } { \color{#FF6800}{ 1 } }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = 8 , \alpha \beta = \color{#FF6800}{ 3 }$
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