# Calculator search results

Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = \dfrac { \left ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } - 2 x \left ( x - 1 \right )$
$y = 0$
$y$-intercept
$\left ( 0 , - 5 \right )$
Maximum
$\left ( \dfrac { 7 } { 2 } , - \dfrac { 11 } { 12 } \right )$
Standard form
$y = - \dfrac { 1 } { 3 } \left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } - \dfrac { 11 } { 12 }$
$\dfrac{ \left( 5x-4 \right) \left( x+1 \right) -11 }{ 3 } -2x \left( x-1 \right) = 0$
$\begin{array} {l} x = \dfrac { 7 + \sqrt{ 11 } i } { 2 } \\ x = \dfrac { 7 - \sqrt{ 11 } i } { 2 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ \dfrac { \left ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } } - 2 x \left ( x - 1 \right ) = 0$
 Arrange the fraction expression 
$\color{#FF6800}{ \dfrac { 5 x ^ { 2 } + x - 15 } { 3 } } - 2 x \left ( x - 1 \right ) = 0$
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
 Expand the expression 
$\dfrac { 5 x ^ { 2 } + x - 15 } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ \dfrac { 5 x ^ { 2 } + x - 15 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 0$
 Calculate the expression as a fraction format 
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - 7 x + 15 } { 3 } } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - 7 x + 15 } { 3 } } = \color{#FF6800}{ 0 }$
 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 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = - 15 + \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
 When raising a fraction to the power, raise the numerator and denominator each to the power 
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = - 15 + \dfrac { \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 ^ { 2 } } { 2 ^ { 2 } } }$
 Organize the expression 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 4 } }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 4 } } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 11 } i } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { \sqrt{ 11 } i } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 11 } i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 11 } i } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 11 } i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 11 } i } { 2 } } \end{array}$
 Organize the expression 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 + \sqrt{ 11 } i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 - \sqrt{ 11 } i } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { 7 + \sqrt{ 11 } i } { 2 } \\ x = \dfrac { 7 - \sqrt{ 11 } i } { 2 } \end{array}$
Calculate using the quodratic formula$($Imaginary root solution$)$\color{#FF6800}{ \dfrac { \left ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } } - 2 x \left ( x - 1 \right ) = 0 $Arrange the fraction expression$ \color{#FF6800}{ \dfrac { 5 x ^ { 2 } + x - 15 } { 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 { 5 x ^ { 2 } + x - 15 } { 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 { x ^ { 2 } - 7 x + 15 } { 3 } } = 0\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - 7 x + 15 } { 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 { - \left ( - 7 \right ) \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 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 { 7 \pm \sqrt{ 7 ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 } } $Organize the expression$ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 \pm \sqrt{ - 11 } } { 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 { 7 \pm \sqrt{ 11 } i } { 2 } } $Separate the answer$ \begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 + \sqrt{ 11 } i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 - \sqrt{ 11 } i } { 2 } } \end{array} $Do not have the solution$ $Calculate using the quadratic formula$\color{#FF6800}{ \dfrac { \left ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } } - 2 x \left ( x - 1 \right ) = 0 $Arrange the fraction expression$ \color{#FF6800}{ \dfrac { 5 x ^ { 2 } + x - 15 } { 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 { 5 x ^ { 2 } + x - 15 } { 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 { x ^ { 2 } - 7 x + 15 } { 3 } } = 0\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - 7 x + 15 } { 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 }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 { 7 \pm \sqrt{ 7 ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 } } $Organize the expression$ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 \pm \sqrt{ - 11 } } { 2 } }\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 \pm \sqrt{ - 11 } } { 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 ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } } - 2 x \left ( x - 1 \right ) = 0 $Arrange the fraction expression$ \color{#FF6800}{ \dfrac { 5 x ^ { 2 } + x - 15 } { 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 { 5 x ^ { 2 } + x - 15 } { 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 { x ^ { 2 } - 7 x + 15 } { 3 } } = 0\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - 7 x + 15 } { 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 15D = \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 15 $Calculate power$ D = \color{#FF6800}{ 49 } - 4 \times 1 \times 15D = 49 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 15 $Multiplying any number by 1 does not change the value$ D = 49 - 4 \times 15D = 49 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 15 } $Multiply$ - 4 $and$ 15D = 49 \color{#FF6800}{ - } \color{#FF6800}{ 60 }D = \color{#FF6800}{ 49 } \color{#FF6800}{ - } \color{#FF6800}{ 60 } $Subtract$ 60 $from$ 49D = \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 ( 5 x - 4 \right ) \left ( x + 1 \right ) - 11 } { 3 } } - 2 x \left ( x - 1 \right ) = 0 $Arrange the fraction expression$ \color{#FF6800}{ \dfrac { 5 x ^ { 2 } + x - 15 } { 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 { 5 x ^ { 2 } + x - 15 } { 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 { x ^ { 2 } - 7 x + 15 } { 3 } } = 0\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x ^ { 2 } - 7 x + 15 } { 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 { - 7 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 15 } { 1 } }\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 7 } { 1 } } , \alpha \beta = \dfrac { 15 } { 1 } $Solve the sign of a fraction with a negative sign$ \alpha + \beta = \color{#FF6800}{ \dfrac { 7 } { 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 } $그래프 보기$ \$
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