All categories

3 years ago

What are the values of x in this equation?

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

4 0

Answer:

A. x = 0, x = -3, x = 2, and x = -2

Step-by-step explanation:

$x^{4} +12x=4x^{2} +3x^{3}$

by moving the terms on the left side to the right side

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

factor an $x^{3}$ from $x^{4} -3x^{3}$

and factor - 4 $x$ from $-4x^{2} +12x$

the equation becomes :

$x^{3} (x - 3 )-4x (x - 3)$ = 0

take $(x - 3)$ as a common factor the equation becomes :

$(x-3)(x^{3}-4x )$ = 0

thus

$x - 3 = 0$ ⇒ $x = 3$

and

$x^{3} - 4x = 0$ by factoring an x we get :

$x(x^{2}-4 ) =0$ ⇒ $x$ = 0 and $x^{2} - 4 = 0$

$x^{2} - 4 =0$ ⇒ $x = 2$ and $x = -2$

Thus

the values are :

x = 0, x = -3, x = 2, and x = -2

You might be interested in

What are the Solutions to the quadratic equation 40-x^2=0

chubhunter [2.5K]

Answer:

the answer will be number two x=+4 $\sqrt 20$

Step-by-step explanation:

if u can brainliest plsss

8 0

3 years ago

Read 2 more answers

In the Isoceles triangle, the legs

aalyn [17]

Check the picture below.

since we know that x = 3/7, we can then plug that on either of the twin legs, since it's an isosceles, and get the length of each leg, so say let's plug it in 5x + 16

$5x + 16\implies 5\left( \cfrac{3}{7} \right)+16\implies \cfrac{38}{7}+16\implies \cfrac{38+112}{7}\implies \cfrac{150}{7}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\\stackrel{\textit{both legs}}{\cfrac{150}{7}+\cfrac{150}{7}}+\stackrel{\textit{base}}{6}\implies \cfrac{150+150+42}{7}\implies \cfrac{342}{7}\implies 48\frac{6}{7}\impliedby \textit{perimeter}$

3 0

3 years ago

If x= -3, and -3 = y, then what can be said about the relationship between x and y? Explain

ella [17]

I'd say that the relationship between y and x is that they're both equivalent to each other, since -3 = -3.

7 0

3 years ago

Read 2 more answers

4) Durante un día de enero la temperatura bajó de 5º a -5º. ¿Cuál fue el cambio en temperatura?

mezya [45]

Temperature goes down up to 10° Celsius during the night.

Given that;

Temperature drop from 5º Celsius to -5º Celsius

Find:

Change in temperature in Celsius

Computation:

Change in temperature = Initial temperature in Celsius + Final temperature in Celsius

Change in temperature = 5 - (-5)

Change in temperature = 5 + 5

Change in temperature = 10° Celsius

Learn more:

brainly.com/question/17429019?referrer=searchResults

4 0

1 year ago

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

2 years ago