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posledela
3 years ago
8

Does the parabola with equation x2 – x + 4 = 0 have real or imaginary roots?

Mathematics
2 answers:
Harman [31]3 years ago
8 0
4. If the determinant delta of a quadratic equation is positive, the equation has two real roots. If the determinant is negative, it has two imaginary roots. In this case, delta=b^2-4ac=(-1)^2-4*1*4=-15. Therefore, this equation has two imaginary roots.
adoni [48]3 years ago
6 0

Answer:

The parabola with equation x^2\:-\:x\:+\:4\:=\:0 has two imaginary roots, because the discriminant is negative.

Step-by-step explanation:

The quadratic formula says that the solutions are

                                                    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

for any quadratic equation like:

                                                   ax^2+bx+c=0

The discriminant is the part of the quadratic formula under the square root.

The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation.

  • A positive discriminant indicates that the quadratic has two distinct real number solutions.
  • A discriminant of zero indicates that the quadratic has a repeated real number solution.
  • A negative discriminant indicates that neither of the solutions are real numbers but there are two imaginary roots that are complex conjugates.

We have the parabola with equation x^2\:-\:x\:+\:4\:=\:0

\mathrm{For\:}\quad a=1,\:b=-1,\:c=4:\quad x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}

The discriminant for this equation is \sqrt{\left(-1\right)^2-4\cdot \:\:1\cdot \:\:4}= \sqrt{-15}, because the discriminant is negative the parabola has two imaginary roots.

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m=7/2

Step-by-step explanation:

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Answer:

We conclude that (4, 2) is NOT a solution to the system of equations.

Step-by-step explanation:

Given the system of equations

y = x - 2

y = 3x + 4

Important Tip:

  • In order to determine whether (4, 2) is a solution to the system of equations or not, we need to solve the system of equations.

Let us solve the system of equations using the elimination method.

\begin{bmatrix}y=x-2\\ y=3x+4\end{bmatrix}

Arrange equation variables for elimination

\begin{bmatrix}y-x=-2\\ y-3x=4\end{bmatrix}

Subtract the equations

y-3x=4

-

\underline{y-x=-2}

-2x=6

Now, solve -2x = 6 for x

-2x=6

Divide both sides by -2

\frac{-2x}{-2}=\frac{6}{-2}

Simplify

x=-3

For y - x = -2 plug in x = -3

y-\left(-3\right)=-2

y+3=-2

Subtract 3 from both sides

y+3-3=-2-3

Simplify

y=-5

The solution to the system of equations is:

(x, y) = (-3, -5)

Checking the graph

From the graph, it is also clear that (4, 2) is NOT a solution to the system of equations because (-3, -5) is the only solution as we have found earlier.

Therefore, we conclude that (4, 2) is NOT a solution to the system of equations.

8 0
3 years ago
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