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Makovka662 [10]
3 years ago
11

Da A Solve the equation 2x^2+3=0

Mathematics
1 answer:
Anika [276]3 years ago
4 0

Answer:

x = ±i(√6 / 2)

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

<u>Algebra I</u>

  • Solving quadratics

<u>Algebra II</u>

Imaginary root i

  • √-1 = i

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

2x² + 3 = 0

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. [Subtraction Property of Equality] Subtract 3 on both sides:                        2x² = -3
  2. [Division Property of Equality] Divide 2 on both sides:                                  x² = -3/2
  3. [Equality Property] Square root both sides:                                                    x = ±√(-3/2)
  4. Simplify:                                                                                                             x = ±i(√6 / 2)
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Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.
Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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