x2 - 4x + 4 or (x - 2)2
It is an equation which is insoluble in its domain. However, it may be soluble in a bigger domain.For example, x2 = 2 has no solution in the domain of rational numbers but it does in the real numbers, orx2 = -2 has no solution in the domain of real number but it does in imaginary numbers.
I don't think there is any easy way to estimate it; just use the quadratic equation to calculate the solutions. You can round some of the numbers to get an estimate; in this case you might even do most of the calculation in your head, but it's probably easier just to do the full calculation.
It is an equation with no solutions [in the given domain]. There may (or may not) be solutions if you change the domain.For example, if X is an integer, then 5X = 2 has no solution. But if you change the domain to rational numbers, then X = 2/5 or 0.4 is a solution.
Let's say you have the quadratic equation x2 - 7x + 12 = 0. Plot the graph of y = x2 - 7x + 12. Where y = 0 (when the graph crosses the x-axis) is a solution to the equation. In this case, it crosses at the points (3,0) & (4,0) so the solutions are x = 3 and x = 4. Now if the graph never touches the x-axis, that means the solutions to the equation are complex numbers.
A quadratic equation always has TWO (2) solutions. They may be different, the same, or non-existant as real numbers (ie they only exist as complex numbers).
0x2 + 1x - 7 = 0
When you put the number in place of the variable (like maybe the 'x' or the 'y') wherever the variable occurs in the equation, and the statement you get out of all the numbers you have is not true, then you know the number is not a solution.
It is an equation which is insoluble in its domain. However, it may be soluble in a bigger domain.For example, x2 = 2 has no solution in the domain of rational numbers but it does in the real numbers, orx2 = -2 has no solution in the domain of real number but it does in imaginary numbers.
Plug 'a', 'b', and 'c' from the equation into the formula. When you do that, the formula becomes a pair of numbers ... one number when you pick the 'plus' sign, and another number when you pick the 'minus' sign. Those two numbers are the 'solutions' to the quadratic equation you started with.
DIVIDE BY ZERO ERROR Is an equation with no solution's answer. * * * * * It also depends on the domain of the variable(s). For example x + 3 = 2 has no solution if the domain for x is the counting numbers, Z. x*3 = 2 has no solution if the domain for x is the natural numbers, N. x2 = 2 has no solution if the domain for x is the rational numbers, Q. x2 = -2 has no solution if the domain for x is the real numbers, R.
Only that the equation that you are trying to solve is a quadratic, that is to say, the powers of the variable are 2,1 and 0 (or any constant increment of these three numbers). Non-negativity of the discriminant is NOT a condition because you can still use the quadratic formula and get roots that are in the complex domain.
I don't think there is any easy way to estimate it; just use the quadratic equation to calculate the solutions. You can round some of the numbers to get an estimate; in this case you might even do most of the calculation in your head, but it's probably easier just to do the full calculation.
They are page numbers 24 and 25 . ( 24 x 25 = 600 ) The easiest way to solve this is by trial and error. Multiply two consecutive numbers; if the product is too low, try larger numbers, if it is too high, try smaller numbers. You can also write an equation and use the quadratic formula. The equation in this case is x(x+1) = 600. Re-written for use of the quadratic equation, it becomes x2 + x - 600 = 0. This will give you a positive and a negative solution; only the positive solution is sensible in this case.
The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.
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It is an equation with no solutions [in the given domain]. There may (or may not) be solutions if you change the domain.For example, if X is an integer, then 5X = 2 has no solution. But if you change the domain to rational numbers, then X = 2/5 or 0.4 is a solution.
There is no simple way which can be easily applied to all forms of equations. If the variable(s) can be eliminated from the equation then, if what is left isfalse, then the equation has no solution.true, then the equation is an identity.The problem is complicated by the domains under consideration. The following are some examples of equations which have no solutions in one domain but do in another [larger] domain:x + 2 = 0 has no solution in N (natural numbers) but does in Z (integers)x*2 = 1 has no solution in Z but does in Q (rational numbers)X^2 = 2 has no solution in Q but does in R (real numbers)x^2 = - 2 has no solution in R but does in C (complex numbers).