The answer depends on the form in which the quadratic function is given. If it is y = ax2 + bx + c then the x-coordinate of the turning point is -b/(2a)
assuming that this means log5125=x, x=3.
cos20 x cos40 x cos80 = 0.0300 radians = 0.125 degrees (the value for radians is given to four decimal places, the value in degrees is exact)
3x+2 x is a variable. A variable is a symbol (x, y, etc...) that does not have an assigned value.
The problem x = 2 sin x cannot be solved by using algebraic methods. One solution is to draw the graphs of y = x and y = 2 sin x. The two lines will intersect. The values of x where the intersection takes place are the solutions to this problem. You can tell from the graph that one solution is x=0 and verify this contention by noting that 2 sin(0) = 0. You can find the other solution through numerical methods and there are many that will give you the correct solution. Perhaps the simplest is to start with a value of X like pi/2 and then take the average of 2*sin(X) and X. Using that as your new value, again take the average of 2*sin(X) and X. As you continue to do this, the value will get closer and closer to the desired value. After 20 steps or so, the precision of your calculator will probably be reached and you will have a pretty good answer of about 1.89549426703398. (A spreadsheet can be used to make these calculations pretty easily.)
Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a
where ax2 + bx + c = 0 x = [-b +/- √(b2 -4ac)] / 2a
A quadratic involving x and y is usually in the form 'y = ax2 + bx + c'. This form is y in terms of x, so we must rearrange it. y = ax2 + bx + c y/a = x2 + bx/a + c/a y/a = x2 + bx/a + d + e, where c/a = d + e, e = (b/a)2 y/a - e = x2 + bx/a + d y/a -e = (x + b/a)2 √(y/a - e) = x + b/a √(y/a - e) - b/a = x
a = bx + c bx + (c-a) = 0 x = (a-c)/b It's simple because there is only one solution. This is a quadratic equation: ax2 + bx + c = 0 It has two solutions.
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. However, assuming your question to find the roots or solutions of ax2 + bx + c = 0, the answer is x = [-b ± sqrt(b2 - 4ac)]/2a b2 - 4ac is called the discriminant. If the discriminant > 0 then the quadratic equation has two distinct real roots. If the discriminant = 0 then the quadratic equation has one double root. If the discriminant < 0 then the quadratic equation has two distinct complex roots that are conjugates of one another.
In theory you can go down the differentiation route but because it is a quadratic, there is a simpler solution. The general form of a quadratic equation is y = ax2 + bx + c If a > 0 then the quadratic has a minimum If a < 0 then the quadratic has a maximum [and if a = 0 it is not a quadratic!] The maximum or minimum is attained when x = -b/2a and you evaluate y = ax2 + bx + c at this value of x to find the maximum or minimum value of the quadratic.
The coefficient of x
When the quadratic is written in the form: y = ax2 + bx + c then if a > 0 y has a minimum if a < 0 y has a maximum and if a = 0 y is not a quadratic but y = bx + c, and it is linear. The maximum or minimum is at x = -b/(2a)
If the quadratic function is written as ax2 + bx + c, then it has no x-intercepts if the discriminant, (b2 - 4ac), is negative.
A quadratic of the form Ax2 + Bx + C where A=1, B=2 and C=-8
ax3 + bx2 + cx x(ax2 + bx + c) you get one answer as 0.
The general form of a quadratic expression is given as ax2+bx+c where "a" cannot equal zero and "b" is the coefficient of the variable "x" and also the sum of the factors of "c" when "a" is unity. Example: x2+5x+6 = (x+2)(x+3) when factored