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The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.
By the fundamental theorem of algebra, if the monic polynomial equation x 2 + bx + c = 0 has complex coefficients, it must have two (not necessarily distinct) complex roots. Unfortunately, the discriminant b 2 − 4c is not as useful in this situation, because it may be a complex number. Still, a modified version of the general theorem can be ...
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
Thus the fraction 3 / 4 can be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division 3 ÷ 4 (three divided by four). We can also write negative fractions, which represent the opposite of a positive fraction. For example, if 1 / 2 represents a half-dollar profit, then − 1 / 2 represents ...
It is unknown whether these constants are transcendental in general, but Γ( 1 / 3 ) and Γ( 1 / 4 ) were shown to be transcendental by G. V. Chudnovsky. Γ( 1 / 4 ) / 4 √ π has also long been known to be transcendental, and Yuri Nesterenko proved in 1996 that Γ( 1 / 4 ), π, and e π are algebraically independent.
Therefore, the 2y on both sides can be cancelled out, leaving 3 = 6y, or y = 0.5. This is equivalent to subtracting 2y from both sides. At times, cancelling out can introduce limited changes or extra solutions to an equation. For example, given the inequality ab ≥ 3b, it looks like the b on both sides can be cancelled out to give a ≥ 3 as ...
For example, using single-precision IEEE arithmetic, if x = −2 −149, then x/2 underflows to −0, and dividing 1 by this result produces 1/(x/2) = −∞. The exact result −2 150 is too large to represent as a single-precision number, so an infinity of the same sign is used instead to indicate overflow.
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3y 2 = −1 is never solvable, but x 2 − 5y 2 = −1 may be. [27] The first few numbers n for which x 2 − ny 2 = −1 is solvable are