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Most famously, an imaginary number, that is, a number whose square (which is the number multiplied by itself) is negative. All real numbers have positive squares. A complex number, is a number which is the sum of a real number and an imaginary number, and so is also a non-real number.
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Is -4 a nonreal complex number?
No. Negative four is a real number. All real numbers are also complex numbers, so it is a complex number (but it's real, not nonreal)
How many solutions are in x2 plus x plus 4 equals 0?
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis
What are solutions that are not real in algebra?
An example would be a negative number found from using the quadratic formula. That can possibly result from an equation that cannot be set into factored form due to the lack of common factors. Imaginary numbers can be considered nonreal. 2i = 2 times radical -1 The radical looks like the pulse monitor you see in hospitals, fyi.
3x2 - 7x plus 6 equals 0?
Nonreal answer if we put it into the quadratic equation 0=ax^2bx+c x=(-b[+/-](b^2-4ac)^0.5)/2a x=(7[+/-](7^2-4*3*6)^0.5)/(2*3) x=(7[+/-](49-72)^0.5)/6 x=(7[+/-]((-23)^0.5))/6 x=(7[+/-]((-23)^0.5)/6 As you cannot have the square root of a negative number (unless to take iinto account, but lets not go there) the quadratic does not cross the x-axis, so there a no real values that can be found to solve the equation (which i assume you are trying to do).
What are nonreal numbers?
The square of 0 is 0. For any other number, whether it is positive or negative, the square is positive. Thus, the square of any real number is non-negative. This means that only non-negative number can have square roots.So what happens if you are trying to solve the equation x^2 +4 = 0. It is not a particularly complicated equation.Simplify to x^2 = -4 and so x = sqrt(-4). But then what? 2*2 = 4 and (-2)*(-2) is also 4 so neither can be an answer.To get around that, mathematicians introduced imaginary numbers. Then, the answer to the above question is that x = 2i (2j if you are a physicist). And they added the special rule for multiplication: i*i = -1. These are the simplest form of non-real numbers.The next level up are complex numbers and these are numbers of the form a + bi where a and b are real and i is the imaginary number representing the square root of -1.Yet another level up are the quaternions which are numbers of the form a + bi + cj + dk where a, b, c and d are real, and i,j k are similar to the i of the complex numbers but with further multiplication rules for i, j and k. See the link below for more on quaternions.http://en.wikipedia.org/wiki/Quaternion
Is -4 a nonreal complex number?
No. Negative four is a real number. All real numbers are also complex numbers, so it is a complex number (but it's real, not nonreal)
Is one a real complex pure imaginary or nonreal complex number?
One is a complex number and a real number.
How many solutions are in x2 plus x plus 4 equals 0?
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis
A repeating decimal is a real number?
Yes repeating decimals are real numbers. They can fall under the category of rational numbers under real numbers since their repeating decimal patterns allows them to be converted into a fraction. Nonreal numbers are imaginary numbers which are expressed with i, or sqrt(-1).
What are solutions that are not real in algebra?
An example would be a negative number found from using the quadratic formula. That can possibly result from an equation that cannot be set into factored form due to the lack of common factors. Imaginary numbers can be considered nonreal. 2i = 2 times radical -1 The radical looks like the pulse monitor you see in hospitals, fyi.
Laws of signs in math?
Descartes' Rule of Signs is a used for finding the number of zeroes of a polynomial. Descartes' Rule of Signs wont tell you where the polynomial's zeroes are, it just tells you how many there are. The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or less than it by a multiple of 2. Multiple roots of the same value are counted separately. Another way to say this is The number of positive real roots of a polynomial is bounded by the number of changes of sign in its coefficients. Here are some examples to help understand this definition. If f(x) is a linear polynomial (that is, of degree 1), say, f(x) = mx + b then the only root of f(x) is x = -b / m , which is positive only when m and b are of opposite sign Here is an example of the application to a an equation of degree 3. 3x3+5x2-4x+3=0 is a polynomial of degreee 3. The signs, starting at the sign for 3 are (+,+,-,+) where sign change occurs between the the x2 term and the x term and then the x term and the constant term. So there are two sign changes. + to - and - back to + This means there are at most 2 positive real solutions. It general there is that number minus an even number, but in this case that would be 2-2, or 0. Now look at the same exmaple. We want f(-x). To change polynomial f(x) into f(-x), change the signs of all the odd order terms. So f(-x) in your example is -3x3+5x2+4x+3=0 There is a sign change between the 3 and the 5 and then no more. So only one sign change. This changes sign once, so the number of positive roots of f(-x) is one, so there is one negative root of f(x) Now putting both f(x) and f(-x) together, we see there are either zero or two positive roots, either two or zero (respectively) nonreal roots, and exactly one negative root. We know that the nonreal roots always come in pairs and are conjugates. For example if a+bi is a root, so is a-bi. Then if we had 2 real positive roots, we could only have one nonreal root since there are only 3 roots to an equation of degree 3. This is impossible, since we just said nonreal roots come in pairs. So we must have 1 real root and 2 nonreal roots. That tells us the one real root must be negative since we either have 2 or 0 positive roots. If anyone does not fully understand this, please ask and more examples will be given
3x2 - 7x plus 6 equals 0?
Nonreal answer if we put it into the quadratic equation 0=ax^2bx+c x=(-b[+/-](b^2-4ac)^0.5)/2a x=(7[+/-](7^2-4*3*6)^0.5)/(2*3) x=(7[+/-](49-72)^0.5)/6 x=(7[+/-]((-23)^0.5))/6 x=(7[+/-]((-23)^0.5)/6 As you cannot have the square root of a negative number (unless to take iinto account, but lets not go there) the quadratic does not cross the x-axis, so there a no real values that can be found to solve the equation (which i assume you are trying to do).
What are nonreal numbers?
The square of 0 is 0. For any other number, whether it is positive or negative, the square is positive. Thus, the square of any real number is non-negative. This means that only non-negative number can have square roots.So what happens if you are trying to solve the equation x^2 +4 = 0. It is not a particularly complicated equation.Simplify to x^2 = -4 and so x = sqrt(-4). But then what? 2*2 = 4 and (-2)*(-2) is also 4 so neither can be an answer.To get around that, mathematicians introduced imaginary numbers. Then, the answer to the above question is that x = 2i (2j if you are a physicist). And they added the special rule for multiplication: i*i = -1. These are the simplest form of non-real numbers.The next level up are complex numbers and these are numbers of the form a + bi where a and b are real and i is the imaginary number representing the square root of -1.Yet another level up are the quaternions which are numbers of the form a + bi + cj + dk where a, b, c and d are real, and i,j k are similar to the i of the complex numbers but with further multiplication rules for i, j and k. See the link below for more on quaternions.http://en.wikipedia.org/wiki/Quaternion
What are the rules fr multiplication and divisions of integers?
(positive number) x (positive number) = positive number (positive number)/(positive number) = positive number (positive number) x ( negative number) = negative number (positive number)/( negative number) = negative number (negative number) x (negative number) = positive number (negative number)/(negative number) = positive number
What is the formula for the atomic number?
atomic number = number of proton in an element number of proton = number of electron mass number = number of proton + number of neutron therefore... atomic number = mass number - number of neutrons
What does a negative number multiplied by a negative number equal to?
A positive number. Positive Number x Positive Number = Positive Number Positive Number x Negative Number = Negative Number Negative Number x Negative Number = Positive Number
Can a real number that is not a rational number is a?
A real number which is not a rational number is an irrational number.
