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3 years ago

|3x-2|>4 what are the steps to solve for this

Mathematics

1 answer:

Alex73 [517]3 years ago

4 0

bearing in mind that an absolute value expression is in effect a piece-wise function with a ± versions.

$\bf |3x-2|>4\implies \begin{cases} +(3x-2)>4\\ -(3x-2)>4 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(3x-2)>4\implies 3x-2>4\implies 3x>6\implies x>\cfrac{6}{3}\implies \boxed{x>2} \\\\[-0.35em] ~\dotfill\\\\ -(3x-2)>4\implies 3x-2\stackrel{\stackrel{flip}{\downarrow} }{$

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3 years ago

Read 2 more answers

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

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

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago

Need help ASAP !!!!!!!

Anestetic [448]

First we distribute $\frac{1}{5}$ into $n$ and $-\frac{1}{7}$ .

$\frac{1}{5} * n = \frac{1}{5} n$
$\frac{1}{5} * -\frac{1}{7} = - \frac{1}{35}$

That makes the left side of the equation

$\frac{1}{5} n - \frac{1}{35} = \frac{1}{6} n$

Now we subtract $\frac{1}{5}n$ from $\frac{1}{6} n$ , which makes $\frac{1}{30}$

Now the equation is

$-\frac{1}{35} = -\frac{1}{30n}$

Our goal is to isolate $n$ , so we both sides by 30

$30* -\frac{1}{35} = - \frac{1}{30}n * 30$

That will make the equation

$- \frac{30}{35} = -n$

We multiply -1 on both sides of the equation:

$\frac{30}{35} = n$

Finally we simplify $\frac{30}{35}$ , which makes $\frac{6}{7}$

So the answer is $n = \frac{6}{7}$