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3 years ago

Solve 2x^2 +16x+50=0

1 answer:

5 0

$2x^2 +16x+50=0\ \ \ \ \ |:2\\\\x^2+8x+25=0\\\\x^2+2\cdot x\cdot4+25=0\ \ \ \ |+4^2\\\\\underbrace{x^2+2\cdot x\cdot4+4^2}_{(a+b)^2=a^2+2ab+b^2}+25=4^2\ \ \ \ \ |-25\\\\(x+4)^2=16-25\\\\(x+4)^2=-9 < 0-NO\ REAL\ SOLUTION$

$COMPLEX\ SOLUTIONS\\\\(x+4)^2=-9\to x+4=\pm\sqrt{-9}\\\\x+4=-3i\ \vee\ x+4=3i\ \ \ \ |-4\\\\x=-4-3i\ \vee\ x=-4+3i$

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3 years ago

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Mandarinka [93]

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Step-by-step explanation:

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Solution

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1 year ago

N is an integer.<br> write the values of n such that -15<3n<img src="https://tex.z-dn.net/?f=%5Cleq" id="TexFormula1" title="

Hoochie [10]

Answer:

$-4, -3, -2, -1, 0, 1, 2$

Step-by-step explanation:

$\frac{-15}{3}$ (Divide all sides by 3 to make it easier to understand)

$-5$

Now you can just look at all of the possible values for n, knowing that it's an integer.

$-5$ is less than $n$ , so we know that $n$ 's lowest value must be $-4$ .

We also know that $n$ is less than or equal to $2$ , so the highest value of $n$ must be $2$ .

Now you can count through the remaining integers from $-4$ to $2$ .

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3 years ago

Solve the system of equations by using the inverse of the coefficient matrix of the equivalent matrix equation.

Alinara [238K]

The solution for the given system of equations x + 8y = -37, 4x + 8y = -52 is $\left[\begin{array}{ccc}&5&\\\\&4&\end{array}\right]$

Given,

System of equations as,

x + 8y = -37

4x + 8y = -52

We have to solve this by using the inverse of coefficient matrix of the equivalent matrix equation.

That is,

$A=\left[\begin{array}{ccc}a&&b\\\\c&&d\end{array}\right]$

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Now we can solve the equations.

Here we have,

x + 8y = -37

4x + 8y = -52

Now in matrix form,

$\left[\begin{array}{ccc}1&&8\\\\4&&8\end{array}\right]$ $\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right]$ $=\left[\begin{array}{ccc}&-37&\\\\&-52&\end{array}\right]$

A X B

We know that,

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Therefore,

$A^{-1} = \frac{1}{(1X8)-(4X8)} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{8-32} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{-24} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$

Then,

$\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right] =$ $=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$ $\left[\begin{array}{ccc}&\frac{37}{52} &\\\\\end{array}\right]$