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Solve the quadratic equation
Answer
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Number of solution
Answer
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Relationship between roots and coefficients
Answer
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Graph
$y = - x ^ { 2 } + 4$
$y = 0$
$x$Intercept
$\left ( 2 , 0 \right )$, $\left ( - 2 , 0 \right )$
$y$Intercept
$\left ( 0 , 4 \right )$
Maximum
$\left ( 0 , 4 \right )$
Standard form
$y = - x ^ { 2 } + 4$
$-x ^{ 2 } +4 = 0$
$\begin{array} {l} x = 2 \\ x = - 2 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = 0$
$ $ Move the constant to the right side and change the sign $ $
$x ^ { 2 } = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 4 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \pm \sqrt{ \color{#FF6800}{ 4 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 2 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 2 \\ x = - 2 \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 4 \right ) } } { 2 \times 1 }$
$ $ 0 has no sign $ $
$x = \dfrac { \color{#FF6800}{ 0 } \pm \sqrt{ 0 ^ { 2 } - 4 \times 1 \times \left ( - 4 \right ) } } { 2 \times 1 }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 4 \right ) } } { 2 \times 1 }$
$ $ The power of 0 is 0 $ $
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 4 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 0 - 4 \times 1 \times \left ( - 4 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm \sqrt{ 16 } } { 2 \times 1 } }$
$x = \dfrac { 0 \pm \sqrt{ \color{#FF6800}{ 16 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 0 \pm \color{#FF6800}{ 4 } } { 2 \times 1 }$
$x = \dfrac { 0 \pm 4 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 0 \pm 4 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 \pm 4 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 + 4 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 0 - 4 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 0 } + 4 } { 2 } \\ x = \dfrac { 0 - 4 } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = \dfrac { 4 } { 2 } \\ x = \dfrac { 0 - 4 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 4 } { 2 } } \\ x = \dfrac { 0 - 4 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 } { 1 } } \\ x = \dfrac { 0 - 4 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 } { 1 } } \\ x = \dfrac { 0 - 4 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 2 } \\ x = \dfrac { 0 - 4 } { 2 } \end{array}$
$\begin{array} {l} x = 2 \\ x = \dfrac { \color{#FF6800}{ 0 } - 4 } { 2 } \end{array}$
$ $ 0 does not change when you add or subtract $ $
$\begin{array} {l} x = 2 \\ x = \dfrac { - 4 } { 2 } \end{array}$
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ \dfrac { - 4 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ \dfrac { - 2 } { 1 } } \end{array}$
$\begin{array} {l} x = 2 \\ x = \dfrac { - 2 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 4 \right )$
$ $ The power of 0 is 0 $ $
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 4 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 4 \right )$
$ $ 0 does not change when you add or subtract $ $
$D = - 4 \times 1 \times \left ( - 4 \right )$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 4 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = - 4 \times \left ( - 4 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right )$
$ $ Multiply $ - 4 $ and $ - 4$
$D = \color{#FF6800}{ 16 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 16 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 0 , \alpha \beta = - 4$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ Change the symbols of both sides of the equation $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 0 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 4 } { 1 } }$
$\alpha + \beta = - \dfrac { 0 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 4 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = - \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 4 } { 1 }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 4 } { 1 }$
$ $ 0 has no sign $ $
$\alpha + \beta = \color{#FF6800}{ 0 } , \alpha \beta = \dfrac { - 4 } { 1 }$
$\alpha + \beta = 0 , \alpha \beta = \dfrac { - 4 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 0 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
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$4.$ Giải các phương trình trùng $hong$ 
a) $x^{4}-5x^{2}+4=0:$ b) $2x^{4}-3x^{2}-2=0:$ c) $3x^{4}+10x^{2}+3=0$ 
$\bar{5} .0$ Giải các phương $tHnh$ 
a) $\dfrac {\left(x+3\right)\left(x-3\right)} {3}+2=x\left(1-x\right)$ b) $\dfrac {x+2} {x-5}+3=\dfrac {6} {2-x}$ 
c) $\dfrac {4} {x+1}=\dfrac {-x^{2}-x+2} {\left(x+1\right)\left(x+2\right)}$ 
$6.$ Giải các phương $1inb$ 
$a\right)$ $\left(3x^{2}-5x+1\right)\left(x^{2}-4\right)=0:$ b) $\left(2x^{2}+x-4\right)^{2}-\left(2x-1\right)^{2}=0$ 
Luyện tập 
. Giải phương trình trùng phương. 
a) $9x^{4}-10x^{2}+1=0:$ $\tarc{\square } $ $-$ $5x^{4}+2x^{2}-16=10-x^{2}$
10th-13th grade
English
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