That really depends on the situation. Just as there are cases where fractional numbers make sense and others where they don't, and just as there are cases where negative numbers make sense and others where they dont, likewise, there are situations in which complex numbers (with an imaginary part) make sense, and others in which they don't. In many practical cases they don't make sense; in this case, the non-real solution is an indication that there is no practical solution. Example: Use the quadratic formula to determine at what time an object, thrown upwards, will reach a certain height. If you get a complex solution (the discriminant is negative), this means the object will NEVER reach the specified height; it will fall back down before it reaches the desired height.
In this case, the variable "time" can only have real values - imaginary values don't make physical sense. However, please note that there are also situations in which complex numbers DO make sense.
I suggest you use the quadratic formula.
I suggest you use the quadratic formula. In this case, a = 1, b = 5, c = 3.
I suggest you apply the quadratic formula (a = 2, b = -4, c = -1).
In general, there are two steps in solving a given quadratic equation in standard form ax^2 + bx + c = 0. If a = 1, the process is much simpler. The first step is making sure that the equation can be factored? How? In general, it is hard to know in advance if a quadratic equation is factorable. I suggest that you use first the new Diagonal Sum Method to solve the equation. It is fast and convenient and can directly give the 2 roots in the form of 2 fractions. without having to factor the equation. If this method fails, then you can conclude that the equation is not factorable, and consequently, the quadratic formula must be used. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009) The second step is solving the equation by the quadratic formula. This book also introduces a new improved quadratic formula, that is easier to remember by relating the formula to the x-intercepts with the parabola graph of the quadratic function.
There are no integer answers for your question. I suggest using the quadratic equation is is y=b±√b²-4(a*c)∕2*a
The answer is two. Despite its name seems to suggest something to do with four, in a quadratic equation the unknown appears at most to the power of two and so is said to be of second degree. The theorem than pertains here is that the number of roots an equation has is equal to its degrees. However, some of the roots can be repeated - an nth degree equation need not have n different roots. Also the roots do not have to be real. However complex roots ( no real) come in pairs so an equation of odd degree must have at least one real root. A quadratic possibly has no real roots.
The answer will depend on how far the square root sign goes.If you want to solve for "x", I suggest you isolate the square root on the left (if it only covers the "2x" part, move the "1" to the other side of the equation). Then, if you square both sides of the equation, you get a formula which you can easily convert to a form which can be solved with the quadratic equation.
The earliest reference to quadratics was by the ancient Egyptians who used them to assist with land administration and taxation. They were good at recording things though, so there is nothing to suggest an earlier civilization didn't use them but we have no evidence of it.
No. As the term might suggest, it is an acidic solution.
I suggest you use the quadratic formula, with a = 3, b = -2, c = 7.
I suggest you stop worshipping your car!
For satellite your main other option is DirectTV.