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If one of the roots of the quadratic equation is 1 plus 3i what is the other root?
The answer to the question, as stated, is that the other root could be anything. However, if all the coefficients of the quadratic equation are real numbers, then the other root is 1 minus 3i.
What is the square root of 2x plus 1 equals 3?
The answer will depend on how far the square root sign goes.If you want to solve for "x", I suggest you isolate the square root on the left (if it only covers the "2x" part, move the "1" to the other side of the equation). Then, if you square both sides of the equation, you get a formula which you can easily convert to a form which can be solved with the quadratic equation.
What two numbers multiply to give you 30 but add to give you 7?
The two numbers are (1/2)(7 - i root(71)) and (1/2)(7 + i root(71)).Note that this problem has no solution in the real numbers, only the complex solution listed above.
How do you find cube root of 1?
The real cube root of 1 is 1, since 13 = 1. There also a pair of complex cube roots.
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.
