All categories

3 years ago

X^2+4x+20=12x-5 Solve this equation

Mathematics

2 answers:

Scrat [10]3 years ago

8 0

Answer:

4 + (or -) 3i

Step-by-step explanation:

Hi!

Alright, so we're going to move all of the variables and constants to one side. Then, using the quadratic formula, we'll find the answer.

1. Move 12x-5 to the left side of the equation by

a. Subtracting 12x from both sides

b. adding 5 to both sides.

1: x^2+4x+20-12x+5

2. Group the common terms

x^2+4x-12x+20+5 --> x^2-8x+25

3. Look at the problem. We can't factor this.

Explanation for why this isn't factorable (you can skip this)

(1) 25 is positive, so its two factors will both be either positive or negative. We're looking for a negative 8, so let's go with negative.

(2) Factors of 25: 1 & 25, 5 &5

Seeing the problem? Ya, none of the factors of 25 sum to eight. So, instead, we're going to use the quadratic formula.

4. So, the quadratic formula: $\frac{-b+-\sqrt{b^2-4ac} }{2a}$

Our equation is x^2-8x+25=0

a is the constant in front of the x^2

b is the constant in front of the x

c is the term without any x values

so, a=1, b=-8, and c= 25

Into equation:

$\frac{8+-\sqrt{64-4(1)(25)} }{2a}$ = $\frac{8+-\sqrt{64-100} }{2}$ = $\frac{8+-\sqrt{-36} }{2}$

8/2 = 4

sqrt(-36) = 6i

6i/2 = 3i (remember, we still have the two in the denominator)

4 + (or -) 3i

Mekhanik [1.2K]3 years ago

4 0

Answer:

$x=4+ 3\mathbf{i}$

$x=4- 3\mathbf{i}$

Step-by-step explanation:

Solve:

$x^2+4x+20=12x-5$

Move everything to the left side:

$x^2+4x+20-12x+5=0$

Simplify:

$x^2-8x+25=0$

Apply the solver formula:

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

Where a=1, b=-8, c=25. Thus:

$\displaystyle x=\frac{8\pm \sqrt{(-8)^2-4(1)(25)}}{2*1}$

$\displaystyle x=\frac{8\pm \sqrt{64-100}}{2}$

$\displaystyle x=\frac{8\pm \sqrt{-36}}{2}$

The solutions are complex numbers:

$\displaystyle x=\frac{8\pm 6\mathbf{i}}{2}$

Simplifying:

$x=4\pm 3\mathbf{i}$

Two solutions are found:

$x=4+ 3\mathbf{i}$

$x=4- 3\mathbf{i}$

You might be interested in

Write a rational function that matches the given statement:

djyliett [7]

Answer:

i put a graph in there

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Pls help will give brainiest and 5star plss help

saul85 [17]

Answer:

Type in AR, DQ,BS, CP, It's tilted sideways.

Step-by-step explanation:

6 0

3 years ago

Any help needed !<br> I don’t understand:/

tensa zangetsu [6.8K]

Answer:

division i think srry if im wrong

Step-by-step explanation:

6 0

3 years ago

Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =

lara [203]

Answer:

a. The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

b. The solutions are

$\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}$

c. The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}$

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

$A\cdot X=B$

where X is the matrix representing the variables of the system, B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

$X=A^{-1}B$

a. Given the system:

$3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11$

The coefficient matrix is:

$A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}5&17&11\\\end{array}\right]$

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}3&2&4&1&0&0 \\\\ 2&5&3&0&1&0 \\\\ 7&2&2&0&0&1\end{array}\right]$

This matrix can be transformed by a sequence of elementary row operations to the matrix

$\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]$

And the inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]$

Next, multiply $A^ {-1}$ by $B$

$X=A^{-1}\cdot B$

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]$

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

b. To solve this system of equations

$x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12$

The coefficient matrix is:

$A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x&y&z\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}0&12&12\\\end{array}\right]$

The inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]$

The solutions are

$\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}$

c. To solve this system of equations

$4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\$

The coefficient matrix is:

$A=\left[\begin{array}{cccc}4&2&1&5\\3&1&4&7\\2&3&1&6\\3&1&1&3\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]$

The inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]$

The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}$

7 0

3 years ago

Read 2 more answers

Which of the following is equal to 7 1/4

fiasKO [112]

Answer:

29/4

Step-by-step explanation:

convert to improper fraction

4 0

3 years ago