qanda-logo
search-icon
Symbol

Calculator search results

Solve the quadratic equation
Answer
circle-check-icon
expand-arrow-icon
Number of solution
Answer
circle-check-icon
expand-arrow-icon
expand-arrow-icon
Relationship between roots and coefficients
Answer
circle-check-icon
expand-arrow-icon
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = - 2 \sqrt{ 3 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 12 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ a } = \pm \sqrt{ \color{#FF6800}{ 12 } }$
$\color{#FF6800}{ a } = \pm \sqrt{ \color{#FF6800}{ 12 } }$
$ $ Solve a solution to $ a$
$\color{#FF6800}{ a } = \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } }$
$\color{#FF6800}{ a } = \pm \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ a } = \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ a } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = - 2 \sqrt{ 3 } \end{array}$
Calculate using the quadratic formula
$a ^ { 2 } = \color{#FF6800}{ 12 }$
$ $ Move the expression to the left side and change the symbol $ $
$a ^ { 2 } - 12 = 0$
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$a = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
$ $ 0 has no sign $ $
$a = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
$a = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
$ $ The power of 0 is 0 $ $
$a = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 12 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ \color{#FF6800}{ 0 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ \color{#FF6800}{ 48 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$
$a = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 48 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$a = \dfrac { 0 \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { 2 \times 1 }$
$a = \dfrac { 0 \pm 4 \sqrt{ 3 } } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$a = \dfrac { 0 \pm 4 \sqrt{ 3 } } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 0 } \pm \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 2 } } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 0 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 0 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 2 } } } \end{array}$
$\begin{array} {l} a = \dfrac { \color{#FF6800}{ 0 } + 4 \sqrt{ 3 } } { 2 } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} a = \dfrac { 4 \sqrt{ 3 } } { 2 } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 2 } } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} a = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 1 } } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = \dfrac { 2 \sqrt{ 3 } } { \color{#FF6800}{ 1 } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} a = \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \\ a = \dfrac { 0 - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \dfrac { \color{#FF6800}{ 0 } - 4 \sqrt{ 3 } } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \dfrac { - 4 \sqrt{ 3 } } { 2 } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 2 } } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } } { \color{#FF6800}{ 1 } } } \end{array}$
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \dfrac { - 2 \sqrt{ 3 } } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} a = 2 \sqrt{ 3 } \\ a = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 3 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$a ^ { 2 } = \color{#FF6800}{ 12 }$
$ $ Move the expression to the left side and change the symbol $ $
$a ^ { 2 } - 12 = 0$
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 12 \right )$
$ $ The power of 0 is 0 $ $
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 12 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 12 \right )$
$ $ 0 does not change when you add or subtract $ $
$D = - 4 \times 1 \times \left ( - 12 \right )$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 12 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = - 4 \times \left ( - 12 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right )$
$ $ Multiply $ - 4 $ and $ - 12$
$D = \color{#FF6800}{ 48 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 48 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 0 , \alpha \beta = - 12$
Find the sum and product of the two roots of the quadratic equation
$a ^ { 2 } = \color{#FF6800}{ 12 }$
$ $ Move the expression to the left side and change the symbol $ $
$a ^ { 2 } - 12 = 0$
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 0 } } { \color{#FF6800}{ 1 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 12 } } { \color{#FF6800}{ 1 } } }$
$\alpha + \beta = - \dfrac { 0 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 12 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 1 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 1 }$
$ $ 0 has no sign $ $
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 12 } { 1 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { - 12 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
Solution search results
Have you found the solution you wanted?
Try again
Try more features at Qanda!
check-iconSearch by problem image
check-iconAsk 1:1 question to TOP class teachers
check-iconAI recommend problems and video lecture