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Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = x ^ { 2 } - 5 x$
$y = 0$
$x$Intercept
$\left ( 0 , 0 \right )$, $\left ( 5 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Minimum
$\left ( \dfrac { 5 } { 2 } , - \dfrac { 25 } { 4 } \right )$
Standard form
$y = \left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } - \dfrac { 25 } { 4 }$
$x ^{ 2 } -5x = 0$
$\begin{array} {l} x = 0 \\ x = 5 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } = 0$
$ax^{2} + bx = x\left(ax+b\right)$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = 0$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 0 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
 When raising a fraction to the power, raise the numerator and denominator each to the power 
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 5 ^ { 2 } } { 2 ^ { 2 } } }$
 Organize the expression 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 25 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 25 } { 4 } }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 25 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 25 } { 4 } } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{array}$
 Organize the expression 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 0 \end{array}$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } = 0$
 Bind the expressions with the common factor $x$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = 0$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 5 \right ) \pm \sqrt{ \left ( - 5 \right ) ^ { 2 } - 4 \times 1 \times 0 } } { 2 \times 1 }$
 Simplify Minus 
$x = \dfrac { 5 \pm \sqrt{ \left ( - 5 \right ) ^ { 2 } - 4 \times 1 \times 0 } } { 2 \times 1 }$
$x = \dfrac { 5 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 0 } } { 2 \times 1 }$
 Remove negative signs because negative numbers raised to even powers are positive 
$x = \dfrac { 5 \pm \sqrt{ 5 ^ { 2 } - 4 \times 1 \times 0 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 \pm \sqrt{ 5 ^ { 2 } - 4 \times 1 \times 0 } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 \pm \sqrt{ 25 } } { 2 \times 1 } }$
$x = \dfrac { 5 \pm \sqrt{ \color{#FF6800}{ 25 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$x = \dfrac { 5 \pm \color{#FF6800}{ 5 } } { 2 \times 1 }$
$x = \dfrac { 5 \pm 5 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { 5 \pm 5 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 \pm 5 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 + 5 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 - 5 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } } { 2 } \\ x = \dfrac { 5 - 5 } { 2 } \end{array}$
 Add $5$ and $5$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { 5 - 5 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 10 } { 2 } } \\ x = \dfrac { 5 - 5 } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 5 - 5 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 5 - 5 } { 2 } \end{array}$
 Reduce the fraction to the lowest term 
$\begin{array} {l} x = \color{#FF6800}{ 5 } \\ x = \dfrac { 5 - 5 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { 2 } \end{array}$
 Remove the two numbers if the values are the same and the signs are different 
$\begin{array} {l} x = 5 \\ x = \dfrac { 0 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 0 } { 2 } } \end{array}$
 If the numerator is 0, it is equal to 0 
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ 0 } \end{array}$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 0$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 5 ^ { 2 } - 4 \times 1 \times 0$
$D = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 0$
 Calculate power 
$D = \color{#FF6800}{ 25 } - 4 \times 1 \times 0$
$D = 25 - 4 \times 1 \color{#FF6800}{ \times } \color{#FF6800}{ 0 }$
 If you multiply a number by 0, it becomes 0 
$D = 25 + \color{#FF6800}{ 0 }$
$D = 25 \color{#FF6800}{ + } \color{#FF6800}{ 0 }$
 0 does not change when you add or subtract 
$D = 25$
$\color{#FF6800}{ D } = \color{#FF6800}{ 25 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = 5 , \alpha \beta = 0$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 5 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 0 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 5 } { 1 } } , \alpha \beta = \dfrac { 0 } { 1 }$
 Solve the sign of a fraction with a negative sign 
$\alpha + \beta = \color{#FF6800}{ \dfrac { 5 } { 1 } } , \alpha \beta = \dfrac { 0 } { 1 }$
$\alpha + \beta = \dfrac { 5 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 0 } { 1 }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = \color{#FF6800}{ 5 } , \alpha \beta = \dfrac { 0 } { 1 }$
$\alpha + \beta = 5 , \alpha \beta = \dfrac { 0 } { \color{#FF6800}{ 1 } }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = 5 , \alpha \beta = \color{#FF6800}{ 0 }$
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