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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.
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.
Could you have a quadratic function with one real root and one complex root?
Provided some of the coefficients and the constant were imaginary (complex) as well, yes. For example, (x + 2)(x - 3+i) has both a real and an imaginary root, and has coefficients that are also both real and imaginary, i.e. 1, -1+i, and -6+2i.
Does every quadratic equation have 2 roots?
No. The quadratic may have what's known as repeated roots, where it only has one root; for example, x2 + 2x + 1 = (x+1)(x+1) = 0 has only one root at x = -1. It always has roots, but can be imaginary roots, also, when no part of the graph intersects the X axis.
What does it mean when the graph of a quadratic function crosses the x axis twice?
When the graph of a quadratic crosses the x-axis twice it means that the quadratic has two real roots. If the graph touches the x-axis at one point the quadratic has 1 repeated root. If the graph does not touch nor cross the x-axis, then the quadratic has no real roots, but it does have 2 complex roots.
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 is the solution set of the quadratic equation 8x2 plus 2x plus 1 0?
There is no equals sign visible but, taking the equation to be8x2 + 2x + 1 = 0,the solutions are the complex conjugate pair, -1.25 ± 0.330719*i where i is the imaginary square root of -1.
Is the square root of 1 a rational number?
Yes, the square root of 1 is 1.
How do you factorise x squared plus x plus 12?
x^2 + x + 12 = 0 You cannot factorize this equation in order to find real roots, but you can factorize it in order to find its complex roots as (x-r1)(x-r2)=0. Solve the equation by using the quadratic formula, find the complex roots, and then you are able to factorize it. x ^2 +x + 12 = 0 x = (-1 + square root of (1 - 48))/2 or x = (-1 - square root of (1 - 48))/2 x = (-1 + square root of (- 47))/2 or x = (-1 - square root of (- 47))/2 By substituting -1 with i^2 we have: x = (-1 + square root of (47i^2))/2 or x = (-1 - square root of (47i^2))/2 x = (-1 + i(square root of 47))/2 or x = (-1 - i(square root of 47))/2 x = -1/2 + i(square root of 47)/2 or x = -1/2 - i(square root of 47)/2 Now you are able to factorize the equation as: [x +(1/2 - ((square root of 47)/2)i))] [x +(1/2 +(square root of 47)/2)i))] Well, let's make one addition to the comment you cannot factor this equation. When you factor an equation you are really factoring it over some field, such as real numbers, or rational numbers. ( look up fields if you are not familiar with them) This can be factored over the complex numbers, In fact, once you find two roots, call them, r1 and r2, you can always factor as (x-r1)(x-r2)=0. In this case the roots are complex, but the same would hold if they were real. Who wrote this comment is right, but I was talking about solving the equation by factoring, without using the quadratic formula. If we factorize the equation as (x - r1)(x - r2), we only presuppose that roots are complex roots, but we can find them only when we solve the equation by using the quadratic formula. So the meaning was that we can't solve this equation by factoring. I appreciate the comment.
How j equals square root of -1?
in complex j2=-1 so j equals square root of -1
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.
