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4 years ago

Which correctly describes the root of the following cubic equation? x^3-3x^2+4x-12=0

Mathematics

2 answers:

aksik [14]4 years ago

7 0

I know that the factors of 12 are 1, 2, 3, 4, 6, and 12
Therefore, the possible roots are 1, 2, 3, 4, 6, and 12

Lina20 [59]4 years ago

7 0

Answer:

The equation has one real root and two complex roots.

Step-by-step explanation:

The given equation is

$x^3-3x^2+4x-12=0$

The above equation is true form x=3, therefore (x-3) is a factor of above equation.

Use long division or synthetic division method to divide the equation by (x-3).

$(x-3)(x^2+4)=0$

Equate each factor equal to zero.

$x-3=0$

$x=3$

Therefore 3 is a real root of the equation.

$x^2+4=0$

$x^2=-4$

$x=\sqrt{-4}$

$x=\pm 2i$

2i and -2i are complex roots of the equation.

Therefore the equation has one real root and two complex roots.

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What is the greatest common factor of 6 and 9

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3 years ago

Read 2 more answers

A) The equilibrium prices P1 and P2 for two goods satisfy the equations:

Allisa [31]

The value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

<h3 />

The equilibrium prices of the three independent commodities using the inverse matrix are P1 = 4, P2 = 7, P3 = 8 respectively.

<h3>What is a matrix?</h3>

A matrix can be defined as a collection of integers(numbers that are either positive or negative) that are organized in rows and columns to construct a rectangular array. The numbers in this matrix system are referred to as elements.

To determine the values of P1 and P2 for the system of equations given by using an inverse matrix, we have:

9P1 + P2 = 51

3P1 +4P2 = 39

Representing the above data in matrix form, we have:

$\left[\begin{array}{cc}9&1\\3&4\\ \end{array}\right] \left[\begin{array}{c} \mathbf{P_1} \\ \mathbf{P_2}\\ \end{array}\right] = \left[\begin{array}{c} \mathbf{51} \\ \mathbf{39} \\ \end{array}\right]$ which is in the form AX = B

In order for us to determine the values of P1 and P2, Let take the inverse of A⁻¹ on both sides of the AX= B, we have:

$\mathbf{{A^{-1} AX} = A^{-1} B}$

X = A⁻¹ B

Let's start by finding A⁻¹;

$\mathbf{A = \left[\begin{array}{cc}9&1\\3&4\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{36-3}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

Now, Let's Find A⁻¹B;

$\mathbf{A^{-1}B = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] \left[\begin{array}{c}51\\39\\ \end{array}\right] }$

$\mathbf{\implies \dfrac{1}{33}\left[\begin{array}{cc}204&-39\\-153&+351\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}\dfrac{165}{33}\\ \\ \dfrac{198}{33}\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}5\\ \\ 6\end{array}\right] }$

$\left[\begin{array}{c}\mathbf{P_1}\\ \mathbf{P_2}\end{array}\right]= \left[\begin{array}{c}5\\ 6 \end{array}\right] }$

Therefore, we can conclude that the value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

To determine the equilibrium prices of the three independent commodities using the inverse matrix, we have:

P₁ + 2P₂ + 3P₃ = 42

2P₁ + P₂ + 4P₃ = 47

P₁ + 3P₂ + 4P₃ = 57

The matrix in AX = B form is computed as:

$\implies\left[ \begin{array}{ccc}1&2&3\\2&1&4\\1&3&4\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3 \end{array}\right] = \left[\begin{array}{c}42\\47\\ 57\end{array}\right]$