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Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = x ^ { 2 } + 2 \sqrt{ 3 } x + 5$
$y = 0$
$y$Intercept
$\left ( 0 , 5 \right )$
Minimum
$\left ( - \sqrt{ 3 } , 2 \right )$
Standard form
$y = \left ( x + \sqrt{ 3 } \right ) ^ { 2 } + 2$
$x ^{ 2 } +2 \sqrt{ 3 } x+5 = 0$
$\begin{array} {l} x = - \sqrt{ 3 } + \sqrt{ 2 } i \\ x = - \sqrt{ 3 } - \sqrt{ 2 } i \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ - } \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \sqrt{ 3 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ - } \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x + \sqrt{ 3 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ + } \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \sqrt{ 3 } \right ) ^ { 2 } = - 5 + \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } }$
 If you square the radical sign, it will disappear 
$\left ( x + \sqrt{ 3 } \right ) ^ { 2 } = - 5 + \color{#FF6800}{ 3 }$
$\left ( x + \sqrt{ 3 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
 Add $- 5$ and $3$
$\left ( x + \sqrt{ 3 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 3 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 3 } } = \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 2 } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 3 } }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 3 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } \end{array}$
$\begin{array} {l} x = - \sqrt{ 3 } + \sqrt{ 2 } i \\ x = - \sqrt{ 3 } - \sqrt{ 2 } i \end{array}$
Calculate using the quodratic formula$($Imaginary root solution$)$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \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 ( 2 \sqrt{ 3 } \right ) \pm \sqrt{ \left ( 2 \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 } }x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) \pm \sqrt{ \left ( 2 \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 } $Get rid of unnecessary parentheses$ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \pm \sqrt{ \left ( 2 \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 } } { 2 \times 1 } $Solve the power$ x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \left ( \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 } $Calculate power$ x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ 4 } \left ( \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ 4 \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 } } { 2 \times 1 } $If you square the radical sign, it will disappear$ x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ 4 \times \color{#FF6800}{ 3 } - 4 \times 1 \times 5 } } { 2 \times 1 }\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ 4 \times 3 - 4 \times 1 \times 5 } } { 2 \times 1 } } $Organize the expression$ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ - 8 } } { 2 \times 1 } }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ - } \color{#FF6800}{ 8 } } } { 2 \times 1 } $Organize the part that can be taken out of the radical sign inside the square root symbol$ x = \dfrac { - 2 \sqrt{ 3 } \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm 2 \sqrt{ 2 } i } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } $Multiplying any number by 1 does not change the value$ x = \dfrac { - 2 \sqrt{ 3 } \pm 2 \sqrt{ 2 } i } { \color{#FF6800}{ 2 } }\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } \pm 2 \sqrt{ 2 } i } { 2 } } $Separate the answer$ \begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } + 2 \sqrt{ 2 } i } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } - 2 \sqrt{ 2 } i } { 2 } } \end{array}\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } + 2 \sqrt{ 2 } i } { 2 } } \\ x = \dfrac { - 2 \sqrt{ 3 } - 2 \sqrt{ 2 } i } { 2 } \end{array} $Reduce the fraction$ \begin{array} {l} x = \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } \\ x = \dfrac { - 2 \sqrt{ 3 } - 2 \sqrt{ 2 } i } { 2 } \end{array}\begin{array} {l} x = - \sqrt{ 3 } + \sqrt{ 2 } i \\ x = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } - 2 \sqrt{ 2 } i } { 2 } } \end{array} $Reduce the fraction$ \begin{array} {l} x = - \sqrt{ 3 } + \sqrt{ 2 } i \\ x = \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ i } \end{array} $Do not have the solution$ $Calculate using the quadratic formula$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) \pm \sqrt{ \left ( 2 \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 } $Get rid of unnecessary parentheses$ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \pm \sqrt{ \left ( 2 \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 } } { 2 \times 1 } $Solve the power$ x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \left ( \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 } $Calculate power$ x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ \color{#FF6800}{ 4 } \left ( \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 } } { 2 \times 1 }x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ 4 \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 } } { 2 \times 1 } $If you square the radical sign, it will disappear$ x = \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ 4 \times \color{#FF6800}{ 3 } - 4 \times 1 \times 5 } } { 2 \times 1 }\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ 4 \times 3 - 4 \times 1 \times 5 } } { 2 \times 1 } } $Organize the expression$ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ - 8 } } { 2 } }\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 2 \sqrt{ 3 } \pm \sqrt{ - 8 } } { 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}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \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 } = \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 }D = \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 $Solve the power$ D = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5D = \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } \left ( \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5 $Calculate power$ D = \color{#FF6800}{ 4 } \left ( \sqrt{ 3 } \right ) ^ { 2 } - 4 \times 1 \times 5D = 4 \left ( \sqrt{ \color{#FF6800}{ 3 } } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 5 $If you square the radical sign, it will disappear$ D = 4 \times \color{#FF6800}{ 3 } - 4 \times 1 \times 5D = \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } - 4 \times 1 \times 5 $Multiply$ 4 $and$ 3D = \color{#FF6800}{ 12 } - 4 \times 1 \times 5D = 12 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 5 $Multiplying any number by 1 does not change the value$ D = 12 - 4 \times 5D = 12 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } $Multiply$ - 4 $and$ 5D = 12 \color{#FF6800}{ - } \color{#FF6800}{ 20 }D = \color{#FF6800}{ 12 } \color{#FF6800}{ - } \color{#FF6800}{ 20 } $Subtract$ 20 $from$ 12D = \color{#FF6800}{ - } \color{#FF6800}{ 8 }\color{#FF6800}{ D } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } $Since$ D<0 $, there is no real root of the following quadratic equation$  $Do not have the real root$ \alpha + \beta = - 2 \sqrt{ 3 } , \alpha \beta = 5$Find the sum and product of the two roots of the quadratic equation$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \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 { 2 \sqrt{ 3 } } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 5 } { 1 } }\alpha + \beta = - \dfrac { 2 \sqrt{ 3 } } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 5 } { 1 } $If the denominator is 1, the denominator can be removed$ \alpha + \beta = - \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) , \alpha \beta = \dfrac { 5 } { 1 }\alpha + \beta = \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \right ) , \alpha \beta = \dfrac { 5 } { 1 } $Get rid of unnecessary parentheses$ \alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } , \alpha \beta = \dfrac { 5 } { 1 }\alpha + \beta = - 2 \sqrt{ 3 } , \alpha \beta = \dfrac { 5 } { \color{#FF6800}{ 1 } } $If the denominator is 1, the denominator can be removed$ \alpha + \beta = - 2 \sqrt{ 3 } , \alpha \beta = \color{#FF6800}{ 5 }\$
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