To find which has imaginary roots, use the discriminant of the quadratic formula (b2 - 4ac) and see if it's less than 0. (The quadratic formula corresponds to general form of a quadratic equation, y = ax2 + bx + c)A) x2 - 1 = 0= 0 - 4(1)(-1) = 4Therefore, the roots are not imaginary.B) x2 - 2 = 0= 0 - 4(1)(-2) = 8Therefore, the roots are not imaginary.C) x2 + x + 1 = 0= 1 - 4(1)(1) = -3Therefore, the roots are imaginary.D) x2 - x - 1 = 0= 1 - 4(1)(-1) = 5Therefore, the roots are not imaginary.The equation x2 + x + 1 = 0 has imaginary roots.
The roots of the equation [ Ax2 + Bx + C = 0 ] arex = 1/2A [ -B ± sqrt(B2 - 4AC) ]
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
In general, there are two steps in solving a given quadratic equation in standard form ax^2 + bx + c = 0. If a = 1, the process is much simpler. The first step is making sure that the equation can be factored? How? In general, it is hard to know in advance if a quadratic equation is factorable. I suggest that you use first the new Diagonal Sum Method to solve the equation. It is fast and convenient and can directly give the 2 roots in the form of 2 fractions. without having to factor the equation. If this method fails, then you can conclude that the equation is not factorable, and consequently, the quadratic formula must be used. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009) The second step is solving the equation by the quadratic formula. This book also introduces a new improved quadratic formula, that is easier to remember by relating the formula to the x-intercepts with the parabola graph of the quadratic function.
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.
That depends on the equation.
It can tell you three things about the quadratic equation:- 1. That the equation has 2 equal roots when the discriminant is equal to zero. 2. That the equation has 2 distinctive roots when the discriminant is greater than zero. £. That the equation has no real roots when the discriminant is less than zero.
-4,3 are the roots of this equation, so for the values for which the sum of roots is 1 & product is -12
It discriminates between the conditions in which a quadratic equation has 0, 1 or 2 real roots.
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
I suggest you apply the quadratic formula (a = 2, b = -4, c = -1).
Use the quadratic formula, with a = 1, b = -3, c = 2.
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.
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.
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
Oh, dude, for this quadratic equation x^2 - 5x + 4 = 0, the solutions are just the roots of the equation. You can find them by either factoring the quadratic or using the quadratic formula. So, like, the solutions are x = 1 and x = 4. Easy peasy lemon squeezy!
No. Well, it depends what you mean with "any quadratic equation". The quadratic formula can solve any equation that can be converted to the form: ax2 + bx + c = 0 Note that it involves only a single variable. There are other limitations as well; for example, no additional operations. If a variable, or the square of a variable, appears in the denominator (1/x, or 1/x2), then some might say that it is "quadratic", but it might no longer be possible to convert the equation into the standard form named above. Similarly, if you have additional operations such as square roots or higher roots, trigonometric functions, etc., it might not be possible to convert the equation into a form that can be solved by the quadratic formula.