Symbol

Calculator search results

Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = 40 x - 5 x ^ { 2 }$
$y = 75$
$x$Intercept
$\left ( 8 , 0 \right )$, $\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
Maximum
$\left ( 4 , 80 \right )$
Standard form
$y = - 5 \left ( x - 4 \right ) ^ { 2 } + 80$
$40x-5x ^{ 2 } = 75$
$\begin{array} {l} x = 5 \\ x = 3 \end{array}$
Find solution by method of factorization
$40 x - 5 x ^ { 2 } = \color{#FF6800}{ 75 }$
 Move the expression to the left side and change the symbol 
$40 x - 5 x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
 Expand the expression 
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 3 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
 Organize the expression 
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
 Move the expression to the left side and change the symbol 
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 Divide both sides by the coefficient of the leading highest term 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
 Organize the expression 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
 Organize the expression 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 3 \end{array}$
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
 Organize the expression 
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
 Move the expression to the left side and change the symbol 
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = 0$
 Bind the expressions with the common factor $5$
$\color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = 0$
$\color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \right ) = \color{#FF6800}{ 0 }$
 Divide both sides by $5$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 8 \right ) \pm \sqrt{ \left ( - 8 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
 Simplify Minus 
$x = \dfrac { 8 \pm \sqrt{ \left ( - 8 \right ) ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$x = \dfrac { 8 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 8 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 15 } } { 2 \times 1 }$
 Remove negative signs because negative numbers raised to even powers are positive 
$x = \dfrac { 8 \pm \sqrt{ 8 ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 \pm \sqrt{ 8 ^ { 2 } - 4 \times 1 \times 15 } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 \pm \sqrt{ 4 } } { 2 \times 1 } }$
$x = \dfrac { 8 \pm \sqrt{ \color{#FF6800}{ 4 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$x = \dfrac { 8 \pm \color{#FF6800}{ 2 } } { 2 \times 1 }$
$x = \dfrac { 8 \pm 2 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { 8 \pm 2 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 \pm 2 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 + 2 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 8 - 2 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 8 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } } { 2 } \\ x = \dfrac { 8 - 2 } { 2 } \end{array}$
 Add $8$ and $2$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { 8 - 2 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 10 } { 2 } } \\ x = \dfrac { 8 - 2 } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 8 - 2 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 8 - 2 } { 2 } \end{array}$
 Reduce the fraction to the lowest term 
$\begin{array} {l} x = \color{#FF6800}{ 5 } \\ x = \dfrac { 8 - 2 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 8 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 2 } \end{array}$
 Subtract $2$ from $8$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 6 } } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 6 } { 2 } } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 3 } { 1 } } \end{array}$
 Reduce the fraction to the lowest term 
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ 3 } \end{array}$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
 Organize the expression 
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
 Move the expression to the left side and change the symbol 
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 75 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 40 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 75$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 40 ^ { 2 } - 4 \times 5 \times 75$
$D = \color{#FF6800}{ 40 } ^ { \color{#FF6800}{ 2 } } - 4 \times 5 \times 75$
 Calculate power 
$D = \color{#FF6800}{ 1600 } - 4 \times 5 \times 75$
$D = 1600 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 75 }$
 Multiply the numbers 
$D = 1600 \color{#FF6800}{ - } \color{#FF6800}{ 1500 }$
$D = \color{#FF6800}{ 1600 } \color{#FF6800}{ - } \color{#FF6800}{ 1500 }$
 Subtract $1500$ from $1600$
$D = \color{#FF6800}{ 100 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = 8 , \alpha \beta = 15$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } = 75$
 Organize the expression 
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } = 75$
$- 5 x ^ { 2 } + 40 x = \color{#FF6800}{ 75 }$
 Move the expression to the left side and change the symbol 
$- 5 x ^ { 2 } + 40 x \color{#FF6800}{ - } \color{#FF6800}{ 75 } = 0$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 40 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 75 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 40 } { 5 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 75 } { 5 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 40 } { 5 } } , \alpha \beta = \dfrac { 75 } { 5 }$
 Solve the sign of a fraction with a negative sign 
$\alpha + \beta = \color{#FF6800}{ \dfrac { 40 } { 5 } } , \alpha \beta = \dfrac { 75 } { 5 }$
$\alpha + \beta = \color{#FF6800}{ \dfrac { 40 } { 5 } } , \alpha \beta = \dfrac { 75 } { 5 }$
 Reduce the fraction 
$\alpha + \beta = \color{#FF6800}{ 8 } , \alpha \beta = \dfrac { 75 } { 5 }$
$\alpha + \beta = 8 , \alpha \beta = \color{#FF6800}{ \dfrac { 75 } { 5 } }$
 Reduce the fraction 
$\alpha + \beta = 8 , \alpha \beta = \color{#FF6800}{ 15 }$
Have you found the solution you wanted?
Try again
Try more features at Qanda!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture