Suppose the revenue equation is of the form R = ax2 + bx + c where a, b and c are constants and x is the variable.
To have a maximum, either a must be negative or x must lie within fixed limits.
If a is negative then the maximum revenue is attained when x = -b/(2a). That is, find the value of R when x = -b/(2a).
If a is positive, then find the value of R when x is at each end point of its domain. One of them will be larger and that is the maximum value of the revenue.
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Yes, however not all quadratic equations can easily be solved by factoring, sometimes you can factor and sometimes it is easier to use the quadratic formula. Example: x2 + 4x + 4 This can be easily factored to (x + 2)(x +2) Therefore the answer is -2 by setting x +2 = 0 and solving for x This can be done using the quadratic equation and you would get the same results, however, it was much faster to factor instead.
( x - 2)( x - 6) = 0 x = 2 or x = 6
For any quadratic ax2 + bx + c = 0 we can find x by using the quadratic formulae: the quadratic formula is... [-b +- sqrt(b2 - 4(a)(c)) ] / 2a
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Using the quadratic equation formula or completing the square
actoring, using the square roots, completing the square and the quadratic formula.
The quadratic formula is used today to find the solutions to quadratic equations, which are equations of the form ax^2 + bx + c = 0. By using the quadratic formula, we can determine the values of x that satisfy the quadratic equation and represent the points where the graph of the equation intersects the x-axis.
Equations using multiple variables, or powers of variables, may not provide a simple numerical value for a given variable. Equations that are solvable using the quadratic formula may result in two values.
Yes, however not all quadratic equations can easily be solved by factoring, sometimes you can factor and sometimes it is easier to use the quadratic formula. Example: x2 + 4x + 4 This can be easily factored to (x + 2)(x +2) Therefore the answer is -2 by setting x +2 = 0 and solving for x This can be done using the quadratic equation and you would get the same results, however, it was much faster to factor instead.
Completing the square is one method for solving a quadratic equation. A quadratic equation can also be solved by factoring, using the square roots or quadratic formula. Solving quadratic equations by completing the square will always work when solving quadratic equations-You can also use division or even simply take a GCF, set the quantities( ) equal to zero, and subtract or add to solve for the variable
It means you are required to "solve" a quadratic equation by factorising the quadratic equation into two binomial expressions. Solving means to find the value(s) of the variable for which the expression equals zero.
Type your answer here...5x2 - 3x + 10 = 2x2x2 - 6x - 7 = 2x
Start with a quadratic equation in the form � � 2 � � � = 0 ax 2 +bx+c=0, where � a, � b, and � c are constants, and � a is not equal to zero ( � ≠ 0 a =0).
If the two roots are x = r1 and x = r2 then the quadratic equation is: (x - r1)(x - r2) = x2 - (r1 + r2)x + r1r2 = 0
( x - 2)( x - 6) = 0 x = 2 or x = 6