I"m trying mine hand on these types of expressions. How to deal with $x$ in $(x+1)^4+(x-1)^4=16$? you re welcome write any type of idea you have, and shot to store it simple. Thanks.

You are watching: Solve for x: −4|x + 5| = −16

Using the development $$\left(a+b\right)^n = \sum_r=0^n\binomnra^rb^n-r$$

$$\requirecancelx^4+\cancel4x^3+6x^2+\cancel4x+1+x^4+-\cancel4x^3+6x^2-\cancel4x+1 = 16$$

$$x^4+6x^2+1=8$$

$$t^2+6t-7=0\quad\textwith\quad x^2=t$$

$$t=1,-7$$

Since $x^2=-7$ doesn"t have actually solutions in reals, $x^2=1\implies x=\pm1$

We have actually our equation to resolve for:$$(x+1)^4+(x-1)^4=16$$First, broaden the terms in the left hand side.$$x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1=16$$Simplify the left hand side.$$2x^4+12x^2+2=16$$Factor $2$ out from both sides.$$x^4+6x^2+1=8$$Move $8$ to the left hand side.$$x^4+6x^2-7=0$$Let $x^2=a$. Now our equation becomes:$$a^2+6a-7=0$$Factor it.$$(a-1)(a+7)=0$$Reverse the substitution.$$(x^2-1)(x^2+7)=0$$Factor some much more (yes even the amount of squares)$$(x+1)(x-1)(x+\sqrt7i)(x-\sqrt7i)=0$$The four roots room $x=\pm 1$ and also $x=\pm \sqrt7i$. Yet if you do not want facility roots, then the two genuine roots room $x=1$, $x=-1$.

Hope ns helped!

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edited Apr 5 "14 at 17:36
reply Apr 5 "14 in ~ 17:22

TrueDefaultTrueDefault
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Let signify by $P(X)$ the adhering to polynomial:

$$P(X)=(X+1)^4+(X-1)^4-16.\quad\quad(1)$$

We immediately see the $X=1$ and also $X=-1$ space roots the $P(X)$. Therefore, $P(X)$ deserve to be created as:

$$P(X)=(X-1)(X+1)Q(X), \quad\quad\;\;\;\;\;\;(2)$$where $Q(X)$ is a polynomial with degree $2$. Hence, $Q(X)=aX^2+bX+c$ for part $a$, $b$, and $c$.

Now, if we construct $P(X)$ for both $(1)$ and $(2)$ us get:

First, v $(1)$:$$P(X)=2X^4+12X^2-14.\quad\quad(3)$$Second, v $(2)$:$$P(X)=(X-1)(X+1)(aX^2+bX+c)=aX^4+bX^3+(c-a)X^2-bX-c.\quad(4)$$Hence, $(3)=(4)$

$$2X^4+12X^2-14=aX^4+bX^3+(c-a)X^2-bX-c.$$

Then, $$\beginarraylr a=2\\ b=0\\ c=14 \endarray$$

And finally,

$$P(X)=(X+1)(X-1)(2X^2+14)=2(X+1)(X-1)(X^2+7).$$

Now the is clear just how to settle $P(X)=0$ (how to uncover the two various other roots):

$$S_\naipublishers.combbR=\-1, +1\.$$and$$S_\naipublishers.combbC=\-1, +1, +i\sqrt7, -i\sqrt7\.$$

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answered Apr 5 "14 in ~ 17:53
npisinpnpisinp
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Looking in ~ $(x+1)^4 + (x-1)^4=16$, i recall that $16=2^4$. So i look in ~ the equation trying to watch if I might get a $2^4$ in there. This sugguests that us guess $x=1$ to do the left hatchet a power of $2$.

Plugging in $x=1$,

$$(1+1)^4 + (1-1)^4 = 2^4 + 0^4 = 16$$

So $x=1$ is a solution.

This trouble has a the opposite property. An alert that sending out $x\rightarrow -x$ doesn"t adjust the expression on the left.

$$(x+1)^4 + (x-1)^4 \rightarrow (-x+1)^4 + (-x-1)^4 = (x-1)^4+(x+1)^4$$

So we know that $x=-1$ will additionally be a solution.

Some thought should convince you the only feasible solutions are between $-1$ and $1$ because any large values of $x$ will automatically make the expression on the left bigger than $16$.

Looking in ~ the derivative that the expression top top the left,

$$y" = 4(x+1)^3 + 4(x-1)^3,$$

we watch that the derivative is $0$ at $x=0$ and also monotonically increasing for $x > 0$ which means that there room no various other solutions since our graph deserve to only crossing the horizontal heat $y=16$ once.

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answer Apr 5 "14 in ~ 18:38
SpencerSpencer
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$(x+1)^4+(x-1)^4=16\\\implies (x^4+4x^3+6x^2+4x+1)+(x^4-4x^3+6x^2-4x+1)-16=0\\\implies (2x^4+12x^2+2)-16\\\implies 2x^4+12x^2-14=0\\\implies 2(x^4+6x^2-7)=0\\\implies 2((x^2)^2+6(x^2)^1-7(x^2)^0)=0$

Now permit $x^2=a$:

$2((a)^2+6(a)^1-7(a)^0)=0\\\implies 2(a^2+6a-7)=0\\\implies a^2+6a-7=0\\\implies (a+7)(a-1)=0\\\implies a=-7 \text or a=1$.

Letting $a=x^2$ again, we acquire

$x^2=-7\text or x^2=1$.

Hence the actual solutions that $x$ room $\pm1$.

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answered Apr 5 "14 at 17:18
Sujaan KunalanSujaan Kunalan
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answered Apr 5 "14 at 17:47
Ivan GandacovIvan Gandacov
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reply Apr 5 "14 in ~ 18:18
GhettoGhetto
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