Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
Yes it is possible. The solutions for a quadratic equation are the points where the function's graph touch the x-axis. There could be 2 places to that even if the graph looks different.
No because quadratic equations only have 2 X-Intercepts
To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions. The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions. To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.
The given expression is a quadratic equation. To find its solutions, we can either factor the equation or use the quadratic formula. However, without an equation to solve or any context, it is not possible to provide a numeric answer.
No, it must have two answers.
Yes it is possible. The solutions for a quadratic equation are the points where the function's graph touch the x-axis. There could be 2 places to that even if the graph looks different.
No because quadratic equations only have 2 X-Intercepts
To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions. The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions. To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.
In the graph of a quadratic equation, the plotted points form a parabola. This parabola usually intersects the X axis at two different points. Those two points are also the two solutions for the quadratic equation. Alternatively: Quadratic equations are formed by multiplying two linear equations together. Each of the linear equations has one solution - multiplying two together means that the solution for either is also a solution for the quadratic equation - hence you get two possible solutions for the quadratic unless both linear equations have exactly the same solution. Example: Two linear equations : x - a = 0 x - b = 0 Multiplied together: (x - a) ( x - b ) = 0 Either a or b is a solution to this quadratic equation. Hence most often you have two solutions but never more than two and always at least one solution.
The given expression is a quadratic equation. To find its solutions, we can either factor the equation or use the quadratic formula. However, without an equation to solve or any context, it is not possible to provide a numeric answer.
Is it possible for a quadratic equation to have no real solution? please give an example and explain. Thank you
Yes it is quite possible
No, it must have two answers.
Yes. For example, the equation x2 = 2, which in standard form is x2 - 2 = 0, has the two solutions x = square root of 2, and x = minus square root of 2.
If you mean 2x^2 -3x +8 = 0 then the discriminant works out as -55 which is less than 0 meaning that the equation has no real roots and so therefore no solutions are possible.
When the equation is a polynomial whose highest order (power) is 2. Eg. y= x2 + 2x + 10. Then you can use quadratic formula to solve if factoring is not possible.
yes