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Why do some quadratic equations have solutions which are imaginary or complex numbers?
The solutions of a quadratic equation in the form of ax^2+bx+c=0 are the points at which the parabola of ax^2+bx+c=y touches the x axis. An imaginary or complex solution to such a question implies that the parabola touches the x axis at a point not within the real x-y plane; to represent complex or imaginary answers, one must introduce a third dimension, and then the location at which the parabola crosses the y-axis will be apparent.
What else can I help you with?
A quadratic equation can't have one imaginary solution?
That's true. Complex and pure-imaginary solutions come in 'conjugate' pairs.
Most quadric equations have?
A quadratic equation can have two real solutions, one real solution, or two complex solutions, none of them real.
Why is it better to solve quadratic equations in the complex number system rather than in the real number system?
It is not always better.Although quadratic equations always have solutions in the complex system, complex solutions might not always make any sense. In such circumstances, sticking to the real number system makes more sense that trying to evaluate an impossible solution in the complex field.
How can you find the complex solutions of any quadratic equation?
I suggest you use the quadratic formula.
How do you solve complex cases of quadratic equations?
If the discriminant - the part under the radical sign in the quadratic formula - is negative, then the result is complex, it is as simple as that. You can't convert a complex number to a real number. If a particular problem requires only real-number solutions, then - if the formula gives complex numbers - you can state that there is no solution.
Do quadratic equations always have two real solutions?
No. A quadratic may have two identical real solutions, two different real solutions, ortwo conjugate complex solutions (including pure imaginary).It can't have one real and one complex or imaginary solution.
A quadratic equation can't have one imaginary solution?
That's true. Complex and pure-imaginary solutions come in 'conjugate' pairs.
Most quadric equations have?
A quadratic equation can have two real solutions, one real solution, or two complex solutions, none of them real.
How many solution will there be if the quadratic equation does not touch or cross the x-axis?
0 real solutions. There are other solutions in the complex planes (with i, the imaginary number), but there are no real solutions.
Why is it better to solve quadratic equations in the complex number system rather than in the real number system?
It is not always better.Although quadratic equations always have solutions in the complex system, complex solutions might not always make any sense. In such circumstances, sticking to the real number system makes more sense that trying to evaluate an impossible solution in the complex field.
Can there be a real and imaginary solution to a quadratic equation?
Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Or there can be a pure real solution or there can be a complex solution.For a quadratic equation ax2+ bx + c = 0, it depends on the value of the discriminant [b2 - 4ac], which is the value inside the radical of the quadratic formula.[b2 - 4ac] > 0 : Two distinct real solutions.[b2 - 4ac] = 0 : Two equal real solutions (double root).[b2 - 4ac] < 0 : Two complex solutions; they will be pure imaginary if b = 0, they will have both real and imaginary parts if b is nonzero.
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.
Why is factoring a valuable tool for solving quadratic equations?
In some simple cases, factoring allows you to find solutions to a quadratic equations easily.Factoring works best when the solutions are integers or simple rational numbers. Factoring is useless if the solutions are irrational or complex numbers. With rational numbers which are relatively complicated (large numerators and denominators) factoring may not offer much of an advantage.
How can you find the complex solutions of any quadratic equation?
I suggest you use the quadratic formula.
How do you solve complex cases of quadratic equations?
If the discriminant - the part under the radical sign in the quadratic formula - is negative, then the result is complex, it is as simple as that. You can't convert a complex number to a real number. If a particular problem requires only real-number solutions, then - if the formula gives complex numbers - you can state that there is no solution.
What are the pros and cons of the quadratic equation?
Pros: There are many real life situations in which the relationship between two variables is quadratic rather than linear. So to solve these situations quadratic equations are necessary. There is a simple equation to solve any quadratic equation. Cons: Pupils who are still studying basic mathematics will not be told how to solve quadratic equations in some circumstances - when the solutions lie in the Complex field.
When are these kinds of numbers solutions to quadratic equations?
You need to be more specific. A quadratic equation will have 2 solutions. The 2 solutions can be equal (such as x² + 2x + 1 = 0, solution is -1 and -1). If one of the solutions is a real number, then the other solution will also be a real number. If one of the solutions is a complex number, then the other solution will also be a complex number. [a complex number has a real component and an imaginary component]In the equation: Ax² + Bx + C = 0. The term [B² - 4AC] will determine if the solution is a double-root, or if the answer is real or complex.if B² = 4AC, then a double-root (real).if B² > 4AC, then 2 real rootsif B² < 4AC, then the quadratic formula will produce a square root of a negative number, and the solution will be 2 complex numbers.If B = 0, then the numbers will be either pure imaginary or real, and negatives of each other [ example 2i and -2i are solutions to x² + 4 = 0]Example of 2 real and opposite sign: x² - 4 = 0; 2 and -2 are solutions.
