For example, x = 2 has only one solution and that is x = 2.
but
x2= 4, the squared equation, has two solutions: x = +2 and x = -2
For example, x = 2 has only one solution and that is x = 2.
but
x2= 4, the squared equation, has two solutions: x = +2 and x = -2
For example, x = 2 has only one solution and that is x = 2.
but
x2= 4, the squared equation, has two solutions: x = +2 and x = -2
For example, x = 2 has only one solution and that is x = 2.
but
x2= 4, the squared equation, has two solutions: x = +2 and x = -2
Remains true. But this does not apply to square roots.
An "extraneous solution" is not a characteristic of an equation, but has to do with the methods used to solve it. Typically, if you square both sides of the equation, and solve the resulting equation, you might get additional solutions that are not part of the original equation. Just do this, and check each of the solutions, whether it satisfies the original equation. If one of them doesn't, it is an "extraneous" solution introduced by the squaring.
Yes. Quite often, if you don't, you'll lose solutions. That is, the transformed equation - after taking square roots - will have less solutions than the original equation.
10d+8=12d the origian square is 4x4
1st equation of motion:v=u+at 2nd equation of motion:s=ut+1/2at square 3rd equation of motion:v square-u square=2as
It often helps to square both sides of the equation (or raise to some other power, such as to the power 3, if it's a cubic root).Please note that doing this may introduce additional solutions, which are not part of the original equation. When you square an equation (or raise it to some other power), you need to check whether any solutions you eventually get are also solutions of the original equation.
Easy. Say I was going to do ____ \/ 2 =1.41.... So then you square the left side 2=1.41.... And then you square the other side 2=2 (When you square root a number and you square it, you end up with the original number)
It often helps to isolate the radical, and then square both sides. Beware of extraneous solutions - the new equation may have solutions that are not part of the solutions of the original equation, so you definitely need to check any purported solutions with the original equation.
According to my calculator: 63.61462096 it might help if you give the original equation (unless you are looking for huge decimal answers)
The idea is to use the equation for the area of a triangle, replace the variables you know (in this case, area and base), and solve the resulting equation for that which you don't know (in this case, the height).
square
Brahmagupta gave the solution of the general linear equation in chapter eighteen of Brahmasphutasiddhanta,18.43 The difference between rupas, when inverted and divided by the difference of the unknowns, is the unknown in the equation. The rupas are [subtracted on the side] below that from which the square and the unknown are to be subtracted.[4]Which is a solution equivalent to , where rupasrepresents constants. He further gave two equivalent solutions to the general quadratic equation,18.44. Diminish by the middle [number] the square-root of the rupas multiplied by four times the square and increased by the square of the middle [number]; divide the remainder by twice the square. [The result is] the middle [number].18.45. Whatever is the square-root of the rupasmultiplied by the square [and] increased by the square of half the unknown, diminish that by half the unknown [and] divide [the remainder] by its square. [The result is] the unknown.[4]Which are, respectively, solutions equivalent to,andHe went on to solve systems of simultaneous indeterminate equations stating that the desired variable must first be isolated, and then the equation must be divided by the desired variable's coefficient