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Can a polynomial be no rational zeros but have real zeros?
Yes, a polynomial can have no rational zeros while still having real zeros. This occurs, for example, in the case of a polynomial like (x^2 - 2), which has real zeros ((\sqrt{2}) and (-\sqrt{2})) but no rational zeros. According to the Rational Root Theorem, any rational root must be a factor of the constant term, and if none exist among the possible candidates, the polynomial can still have irrational real roots.
Where do you find complex numbers in algebra?
Certain functions, when solving to find the zeros (value which makes the function equal zero), the only value which will work has an imaginary component. Note that a parabola (graph of a quadratic or 2nd order polynomial) can touch the x-axis at a single point, or 2 points or no points. If it does not touch or cross the x-axis, then the root (or zeros) of the function are complex with imaginary components.Technically, all real numbers are a subset of complex numbers, so all numbers are complex - but this is not how we normally refer to them. We usually say that a number is real, or it is imaginary, or it is complex.
Which number belongs to the set of imaginary numbers 2?
2 does belong to the set of imaginary numbers. Any real number is also imaginary. Imaginary numbers are the set of all numbers that can be expressed as a +b*i where "i" is the square root of negative one and "a" and "b" are both real numbers.
State whether the following is a polynomial function give the zeros both real and imaginary of the function if they exist g x x 2-3x-4 x 2 plus 1?
The function ( g(x) = \frac{x^2 - 3x - 4}{x^2 + 1} ) is not a polynomial function because it is a rational function (the ratio of two polynomials). To find the zeros, we set the numerator equal to zero: ( x^2 - 3x - 4 = 0 ). The zeros can be found using the quadratic formula: ( x = \frac{3 \pm \sqrt{(3)^2 - 4(1)(-4)}}{2(1)} ), which simplifies to ( x = 4 ) and ( x = -1 ). The denominator ( x^2 + 1 = 0 ) gives imaginary zeros ( x = i ) and ( x = -i ).
What are the zeros of polynomial x2-16?
The zeros of the polynomial ( x^2 - 16 ) can be found by setting the equation equal to zero: ( x^2 - 16 = 0 ). This can be factored as ( (x - 4)(x + 4) = 0 ). Therefore, the zeros are ( x = 4 ) and ( x = -4 ).
Find all real and imaginary solutions to each equation?
(3z-2)^2=4
How would you express the zeros of the equation x2 - 2 equals 0 Are the two zeros of this equation integers rational numbers or irrational numbers?
x = sqrt(2). The zeros are irrational.
Is it possible to have a quadratic function with one real zero and one imaginary zero why or why not?
ax2+bx+c=0 with a,b and c real discriminant D=b2-4ac if D >0 then there are 2 real zeros if D = 0 then there is one real zero if D<0 then there are two imaginary zeros There are no other possibility for D For further information search for fundamental theorem of algebra
What happens if there are no zeros in a quadratic function?
Whether or not a function has zeros depends on the domain over which it is defined.For example, the linear equation 2x = 3 has no zeros if the domain is the set of integers (whole numbers) but if you allow rational numbers then x = 1.5 is a zero.A quadratic function such as x^2 = 2 has no rational zeros, but it does have irrational zeros which are sqrt(2) and -sqrt(2).Similarly, a quadratic equation need not have real zeros. It will have zeros if the domain is extended to the complex field.In the coordinate plane, a quadratic without zeros will either be wholly above the horizontal axis or wholly below it.
What is An equation that is true for NO values of the variable.?
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.
What is x2 -2x plus 2 equals 0?
It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.
What are examples of quadratic formula with 2 imaginary roots?
A quadratic equation has the form: x^2 - (sum of the roots)x + (product of the roots) = 0 If the roots are imaginary roots, these roots are complex number a+bi and its conjugate a - bi, where a is the real part and b is the imaginary part of the complex number. Their sum is: a + bi + a - bi = 2a Their product is: (a + bi)(a - bi) = a^2 + b^2 Thus the equation will be in the form: x^2 - 2a(x) + a^2 + b^2 = 0 or, x^2 - 2(real part)x + [(real part)^2 + (imaginary part)^2]= 0 For example if the roots are 3 + 5i and 3 - 5i, the equation will be: x^2 - 2(3)x + 3^2 + 5^2 = 0 x^2 - 6x + 34 = 0 where, a = 1, b = -6, and c = 34. Look at the denominator of this quadratic equation: D = b^2 - 4ac. D = (-6)^2 - (4)(1)(34) = 36 - 136 = -100 D < 0 Since D < 0 this equation has two imaginary roots.
Can a polynomial be no rational zeros but have real zeros?
Yes, a polynomial can have no rational zeros while still having real zeros. This occurs, for example, in the case of a polynomial like (x^2 - 2), which has real zeros ((\sqrt{2}) and (-\sqrt{2})) but no rational zeros. According to the Rational Root Theorem, any rational root must be a factor of the constant term, and if none exist among the possible candidates, the polynomial can still have irrational real roots.
How many real and imaginary solutions does the equation 2x2 -7x plus 9 equals x plus 1 have?
It has two equal solutions for x which are x = 2 and x = 2
If an equation has a degree of three how many solutions will there be?
If the highest degree of an equation is 3, then the equation must have 3 solutions. Solutions can be: 1) 3 real solutions 2) one real and two imaginary solutions.
Where do you find complex numbers in algebra?
Certain functions, when solving to find the zeros (value which makes the function equal zero), the only value which will work has an imaginary component. Note that a parabola (graph of a quadratic or 2nd order polynomial) can touch the x-axis at a single point, or 2 points or no points. If it does not touch or cross the x-axis, then the root (or zeros) of the function are complex with imaginary components.Technically, all real numbers are a subset of complex numbers, so all numbers are complex - but this is not how we normally refer to them. We usually say that a number is real, or it is imaginary, or it is complex.
Which number belongs to the set of imaginary numbers 2?
2 does belong to the set of imaginary numbers. Any real number is also imaginary. Imaginary numbers are the set of all numbers that can be expressed as a +b*i where "i" is the square root of negative one and "a" and "b" are both real numbers.
