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3 years ago

Which is the graph of a quadratic equation that has a negative discriminant?

Mathematics

1 answer:

vovikov84 [41]3 years ago

5 0

Answer:

Here's what I get

Step-by-step explanation:

The formula for a quadratic equation is

ax² + bx + c = 0

The quadratic formula gives the roots:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-b\pm\sqrt{D}}{2a}$

D is the discriminant.

It tells us the number of roots to the equation — the number of times the graph crosses the x-axis.

$D = \begin{cases}\text{positive} & \quad \text{2 real solutions}\\\text{zero} & \quad \text{1 real solution}\\\text{negative} & \quad \text{0 real solutions}\\\end{cases}$

It doesn't matter if the graph opens upwards or downwards.

If D > 0, the graph crosses the x-axis at two points.

If D = 0, the graph touches the x-axis at one point.

If D < 0, the graph never reaches the x-axis.

Your graph must look like one of the two graphs on the right in the Figure below.

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One of the corners of a cube is sliced off. The resulting cross section will have a minimum of how many sides?

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1 and 3, 2 and 4, 5 and 7, and 8 and 65

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2 years ago

What is the solution to the equation?

kow [346]

4x + 6 = 18

-6 is solution ?

4 * (-6) + 6 = 18 ?

-24 + 6 = - 18

so not the solution

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2 years ago

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Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

$2sin(\theta)+1=0$

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

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2 years ago