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IgorLugansk [536]
4 years ago
12

Which of the following statments is true regarding the solutions to the quadratic equation 5x^2 +6x+2=0? See picture

Mathematics
1 answer:
Naddika [18.5K]4 years ago
8 0

Answer:

The quadratic equation has two complex solutions.

Step-by-step explanation:

5x² + 6x + 2 = 0

To know the the correct answer to the question, we shall determine the discriminant of the equation. This can be obtained as follow:

5x² + 6x + 2 = 0

ax² + bx + c = 0

Comparing the above equation, we can obtain:

a = 5

b = 6

c = 2

Discriminant (D) =?

D = b² – 4ac

D = 6² – (4 × 5 × 2)

D = 36 – 40

D = – 4

Since the discriminant of the equation is less than 0, it means that the equation has no real root ( i.e the equation has two complex solution).

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By applying the definition of product between two <em>square</em> matrices, we find that \vec A \,\cdot \,\vec A = \left[\begin{array}{cc}1&2\\3&6\end{array}\right] \cdot \left[\begin{array}{cc}1&2\\3&6\end{array}\right] is equal to the matrix \vec A \,\cdot \,\vec A = \left[\begin{array}{cc}7&14\\21&42\end{array}\right]. (Correct choice: D)

<h3>What is the product of two square matrices</h3>

In this question we must use the definition of product between two <em>square</em> matrices to determine the resulting construction:

\vec A \,\cdot \,\vec A = \left[\begin{array}{cc}1&2\\3&6\end{array}\right] \cdot \left[\begin{array}{cc}1&2\\3&6\end{array}\right]

\vec A \,\cdot \,\vec A = \left[\begin{array}{cc}7&14\\21&42\end{array}\right]

By applying the definition of product between two <em>square</em> matrices, we find that \vec A \,\cdot \,\vec A = \left[\begin{array}{cc}1&2\\3&6\end{array}\right] \cdot \left[\begin{array}{cc}1&2\\3&6\end{array}\right] is equal to the matrix \vec A \,\cdot \,\vec A = \left[\begin{array}{cc}7&14\\21&42\end{array}\right].

To learn more on matrices: brainly.com/question/11367104

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