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3 years ago

Fill in the blanks so that the equation has the solution of x={-4,-1}

Mathematics

1 answer:

alexgriva [62]3 years ago

4 0

Answer:

The equation that gave the solution provided is either

x² + 5x + 4 = 0 or k(x² + 5x + 4) = 0 where k = real number

Step-by-step explanation:

Although, the question isn't complete and the blanks to be filled aren't provided, Hence, we can work out a quadratic equation from the solutions for x given

The solutions for x include x = -4 or x = -1

x = -4 or x = -1

(x + 4) = 0 or (x + 1) = 0

So,

(x+4)(x+1) = 0

x² + 4x + x + 4 = 0

x² + 5x + 4 = 0

Hence, the equation that gave this solution is either x² + 5x + 4 = 0 or a constant multiple of the equation obtained; k(x² + 5x + 4) = 0

Hope this Helps!!!

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Step-by-step explanation:

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2 years ago

Solve the system of equations by finding the reduced row echelon form of the augmented matrix. Write the solutions for x and y i

Gwar [14]

Answer:

The solutions for the given system of equations are:

$\left \{ {{x=-5} \atop {y=12-4z}} \right.$

Step-by-step explanation:

Given the equation system:

$\left \{ {{3x+y+4z=-3} \atop {-x+y+4z=17}} \right.$

We obtain the following matrix:

$\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right]$

Step 1: Multiply the fisrt row by 1/3.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right]$

Step 2: Sum the first row and the second row.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right]$

Step 3: Multiply the second row by 3/4.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&1 &4&12\end{array}\right]$

Step 4: Multiply the second row by -1/3 and sum the the first row.

$\left[\begin{array}{cccc}1&0 &0&-5\\0&1 &4&12\end{array}\right]$

The result of the reduced matrix is:

$\left \{ {{x=-5} \atop {y+4z=12}} \right.$

This is equal to:

$\left \{ {{x=-5} \atop {y=12-4z}} \right.$

These are the solutions for the system of equations in terms of z, where z can be any number.

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