Q: You are about to factor a quadratic equation involving some x s some x's and a plain number you see that the sign in front of the x's is negative and the sign in front of the plain number is positive?

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They each typically have two solutions, a positive one and a negative one.

If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value

If the discriminant is negative, the roots will be two unreal complex conjugates. If the discriminate is positive the roots will be real.

The term inside the square root symbol is called the radicand. There isn't a specific term for it based on its sign; whether it's positive or negative, it's still the radicand.I'm a little confused by your reference to the quadratic equation.If the radicand is negative, the root is an imaginary number, though that doesn't specifically have anything to do with the quadratic equation in particular.If the quantity b2 - 4ac is negative in the quadratic equation, the root of the quadratic equation is either complex or imaginary depending on whether or not b is zero.---------------------------Thank you to whoever answered this first; you saved me a bit of trouble explaining this to the asker :)However, in the quadractic equation, the number under the radical is called the discriminant. This determines the number of solutions of the quadratic. If the radicand is negative, this means that there are no real solutions to the equation.

No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).

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That means that both of your brackets will have minus signs.

They each typically have two solutions, a positive one and a negative one.

That means that both of your brackets will have minus signs.

That means that both of your brackets will have minus signs.

That means that both of your brackets will have minus signs.

That means that both of your brackets will have minus signs.

That means that both of your brackets will have minus signs.

If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value

If the discriminant is negative, the roots will be two unreal complex conjugates. If the discriminate is positive the roots will be real.

The term inside the square root symbol is called the radicand. There isn't a specific term for it based on its sign; whether it's positive or negative, it's still the radicand.I'm a little confused by your reference to the quadratic equation.If the radicand is negative, the root is an imaginary number, though that doesn't specifically have anything to do with the quadratic equation in particular.If the quantity b2 - 4ac is negative in the quadratic equation, the root of the quadratic equation is either complex or imaginary depending on whether or not b is zero.---------------------------Thank you to whoever answered this first; you saved me a bit of trouble explaining this to the asker :)However, in the quadractic equation, the number under the radical is called the discriminant. This determines the number of solutions of the quadratic. If the radicand is negative, this means that there are no real solutions to the equation.

No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).

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