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Q: Are there no real solutions if a quadratic function is 0?

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If the discriminant of a quadratic equation is less then 0 then it will have no real solutions.

b^2 - 4ac, the discriminant will tell you that a quadratic equation may have one real solution( discriminant = 0 ) , two real solutions( discriminant > 0 ), or no real solutions( discriminant < 0 ).

If the discriminant > 0 then 2 distinct real solutions.If the discriminant = 0 then 1 double real solution.If the discriminant < 0 then no real solutions (though there are two complex solutions).

0 real solutions. There are other solutions in the complex planes (with i, the imaginary number), but there are no real solutions.

It's when ax2+bx+c=0 if b2-4ac= is negative

If the discriminant of b2-4ac of the quadratic equation is greater the 0 then it will have 2 solutions.

The discriminant is -439 and so there are no real solutions.

Suppose the quadratic equation is ax^2 + bx + c = 0 and D = b^2 - 4ac is the discriminant. Then the solutions to the quadratic equation are [-b Â± sqrt(d)]/(2a). Since D = 0, the both solutions are equal to -b/(2a), a single real solution.

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.

There are no real solutions because the discriminant of the quadratic equation is less than zero.

The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.

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.

A quadratic equation can have a maximum of 2 solutions. If the discriminant (b2-4ac) turns out to be less than 0, the equation will have no real roots. If the Discriminant is equal to 0, it will have equal roots. But, if the discriminant turns out to be more than 0,then the equation will have unequal and real roots.

They will have 2 different solutions or 2 equal solutions and some times none depending on the value of the discriminant within the quadratic equation

x2 + 49 = 0

The quadratic equation is: Ax2+ Bx + C = 0-- The equation always has two solutions. They arex = 1/2A [ - B + sqrt(B2 - 4AC) ]x = 1/2A [ - B - sqrt(B2 - 4AC) ] .-- The solutions are real if ( B2 > or = 4AC ).-- The solutions are equal if ( B2 = 4AC ).-- The solutions are complex conjugates if ( B2 < 4AC ).-- The solutions are pure imaginary if ( B = 0 ) & (4AC>0 i.e. -4AC

The domain is all real numbers, and the range is nonnegative real numbers (y ≥ 0).

Assuming a, b, and c are real numbers, there are three possibilities for the solutions, depending on whether the discriminant - the square root part in the quadratic formula - is positive, zero, or negative:Two real solutionsOne ("double") real solutionTwo complex solutions

Write the quadratic equation in the standard form: ax2 + bx + c = 0 Then calculate the discriminant = b2 - 4ac If the discriminant is greater than zero, there are two distinct real solutions. If the discriminant is zero, there is one real solution. If the discriminany is less than zero, there are no real solutions (there will be two distinct imaginary solutions).

Yes. It can have 0, 1, or 2 solutions.

You are supposed to calculate the so-called discriminant, b2 - 4ac. If the result is positive, the equation has two real solutions; if it is zero, one real solution; if it is negative, no real solution (and two complex solutions). For this particular equation, a = 1, b = -10, and c = 25.

It depends on the equation. Also, the domain must be such that is supports an infinite number of solutions. A quadratic equation, for example, has no real solution if its discriminant is negative. It cannot have an infinite number of solutions. Many trigonometric equations are periodic and consequently have an infinite number of solutions - provided the domain is also infinite. A function defined as follows: f(x) = 1 if x is real f(x) = 0 if x is not real has no real solutions but an infinite number of solutions in complex numbers.

It has no real roots.

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.

For solving a quadratic, you mean? Well, if you have real solutions you know that your quadratic has real numbers that make f(x) equal 0. For example, the quadratic f(x) = x^2 - x - 6 has -2 and 3 for solutions, because if you put in either of those two numbers for x, you get 0.I don't know how far along in your math career you are, but there aren't really any quadratics with no solutions, just with no real solutions. Some quadratics require you to take the square root of a negative number to get a solution, which is acceptable, it's just not a real solution anymore.