what is the solution set to the following equation?
$x^4 + 2x - 48 = 0$
select one:
○ a. $\{\pm2i\sqrt{2},\pm\sqrt{6}\}$
○ b. $\{\pm i\sqrt{6},\pm\sqrt{2}\}$
○ c. $\{\pm2i\sqrt{2},\pm3\}$
○ d. $\{\pm i\sqrt{3},\pm2\sqrt{2}\}$

question 18
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the solution of the cubic function $f(x)$ are $-4$, $\frac{5}{3}$, and 1.
identify a factor of this function.
select one:
○ a. $3x + 5$
○ b. $5x + 3$
○ c. $x + 1$
○ d. $x + 4$

Question

what is the solution set to the following equation?
$x^4 + 2x - 48 = 0$
select one:
○ a. $\{\pm2i\sqrt{2},\pm\sqrt{6}\}$
○ b. $\{\pm i\sqrt{6},\pm\sqrt{2}\}$
○ c. $\{\pm2i\sqrt{2},\pm3\}$
○ d. $\{\pm i\sqrt{3},\pm2\sqrt{2}\}$

question 18
not yet answered
marked out of 1.00
flag question

the solution of the cubic function $f(x)$ are $-4$, $\frac{5}{3}$, and 1.
identify a factor of this function.
select one:
○ a. $3x + 5$
○ b. $5x + 3$
○ c. $x + 1$
○ d. $x + 4$

Accepted Answer

### First Equation Solution
# Explanation:
## Step1: Test rational root candidates
By Rational Root Theorem, test $x=2$:
$2^4 + 2(2) - 48 = 16 + 4 - 48 = -28
eq 0$
Test $x=3$:
$3^4 + 2(3) - 48 = 81 + 6 - 48 = 39
eq 0$
Test $x=\sqrt{2}$:
$(\sqrt{2})^4 + 2(\sqrt{2}) - 48 = 4 + 2\sqrt{2} - 48 = -44 + 2\sqrt{2}
eq 0$
Test $x=\sqrt{6}$:
$(\sqrt{6})^4 + 2(\sqrt{6}) - 48 = 36 + 2\sqrt{6} - 48 = -12 + 2\sqrt{6}
eq 0$
Rewrite equation as $x^4 = 48 - 2x$. Test $x=2$ fails, test $x=-2$:
$(-2)^4 + 2(-2) - 48 = 16 - 4 - 48 = -36
eq 0$
Test $x=2i\sqrt{2}$:
$(2i\sqrt{2})^4 + 2(2i\sqrt{2}) - 48 = (16 \cdot 4) + 4i\sqrt{2} - 48 = 64 - 48 + 4i\sqrt{2} = 16 + 4i\sqrt{2}
eq 0$
Test $x=\sqrt{6}$:
$(\sqrt{6})^4 + 2\sqrt{6} - 48 = 36 + 2\sqrt{6} - 48 = -12 + 2\sqrt{6}
eq 0$
Test $x=\sqrt{2}$:
$(\sqrt{2})^4 + 2\sqrt{2} - 48 = 4 + 2\sqrt{2} - 48 = -44 + 2\sqrt{2}
eq 0$
Test $x=3$:
$3^4 + 2*3 - 48 = 81 + 6 - 48 = 39
eq 0$
Test $x=i\sqrt{6}$:
$(i\sqrt{6})^4 + 2(i\sqrt{6}) - 48 = (36) + 2i\sqrt{6} - 48 = -12 + 2i\sqrt{6}
eq 0$
Test $x=2i\sqrt{2}$:
$(2i\sqrt{2})^4 = (2^4)(i^4)(\sqrt{2}^4) = 16*1*4=64$
$64 + 2*(2i\sqrt{2}) -48 = 16 +4i\sqrt{2}
eq0$
Test $x=i\sqrt{3}$:
$(i\sqrt{3})^4 +2(i\sqrt{3}) -48 = 9 +2i\sqrt{3}-48=-39+2i\sqrt{3}
eq0$
Test $x=2\sqrt{2}$:
$(2\sqrt{2})^4 +2*(2\sqrt{2}) -48= 64 +4\sqrt{2}-48=16+4\sqrt{2}
eq0$
Wait, factor as quadratic in $x^2$? No, it's $x^4 +0x^3+0x^2+2x-48=0$. Use substitution $y=x^2$, no, linear term exists. Use synthetic division with root $x=2$:
Coefficients: 1,0,0,2,-48
Bring down 1, multiply by 2: 2, add to 0: 2
Multiply 2 by 2:4, add to 0:4
Multiply 4 by2:8, add to2:10
Multiply10 by2:20, add to-48: -28≠0
Root $x=-2$:
Bring down1, multiply by-2:-2, add to0:-2
Multiply-2 by-2:4, add to0:4
Multiply4 by-2:-8, add to2:-6
Multiply-6 by-2:12, add to-48:-36≠0
Root $x=\sqrt{6}$: not rational, test option b: $x=\sqrt{2}$:
$(\sqrt{2})^4 +2\sqrt{2}-48=4+2\sqrt{2}-48=-44+2\sqrt{2}
eq0$
$x=i\sqrt{6}$: $(i\sqrt{6})^4 +2i\sqrt{6}-48= (i^4)(6^2)+2i\sqrt{6}-48=36+2i\sqrt{6}-48=-12+2i\sqrt{6}
eq0$
Option c: $x=3$: $81+6-48=39≠0$
Option a: $x=\sqrt{6}$: $36+2\sqrt{6}-48=-12+2\sqrt{6}≠0$
Wait, correct factorization: $x^4 +2x -48=(x^2 - 8)(x^2 +6)+2x+48-48=(x^2-8)(x^2+6)+2x$ no. Use $x^4 - 36 +2x -12=(x^2-6)(x^2+6)+2(x-6)$ no.
Wait, solve $x^4 =48-2x$. For real roots: $x^4 \geq0$, so $48-2x\geq0\Rightarrow x\leq24$.
Test $x=2$: 16 vs 44, no; $x=3$:81 vs42, no; $x=1$:1 vs46, no; $x=0$:0 vs48, no; $x=-2$:16 vs52, no; $x=-3$:81 vs54, no. So all roots are complex.
Test option b: $x=\sqrt{2}$: $4+2\sqrt{2}-48≠0$; $x=i\sqrt{6}$: $36 +2i\sqrt{6}-48=-12+2i\sqrt{6}≠0$
Option a: $x=2i\sqrt{2}$: $(2i\sqrt{2})^4 +2*(2i\sqrt{2})-48= 16*(4) +4i\sqrt{2}-48=64-48+4i\sqrt{2}=16+4i\sqrt{2}≠0$
Option d: $x=2\sqrt{2}$: $(2\sqrt{2})^4 +2*(2\sqrt{2})-48=64+4\sqrt{2}-48=16+4\sqrt{2}≠0$; $x=i\sqrt{3}$: $(i\sqrt{3})^4 +2i\sqrt{3}-48=9+2i\sqrt{3}-48=-39+2i\sqrt{3}≠0$
Wait, correct approach: Let $x^4 +2x -48=0$
Assume it factors as $(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$
Set equal to $x^4+0x^3+0x^2+2x-48$
So:
1. $a+c=0\Rightarrow c=-a$
2. $-a^2 +b+d=0\Rightarrow b+d=a^2$
3. $a(d-b)=2$
4. $bd=-48$
From 3: $d-b=\frac{2}{a}$
From 2: $b+d=a^2$
Solve: $d=\frac{a^2}{2}+\frac{1}{a}$, $b=\frac{a^2}{2}-\frac{1}{a}$
$bd=(\frac{a^2}{2})^2 - (\frac{1}{a})^2=\frac{a^4}{4}-\frac{1}{a^2}=-48$
Multiply by $4a^2$: $a^6 -4= -192a^2$
$a^6 +192a^2 -4=0$
Let $z=a^2$: $z^3+192z-4=0$
No rational $z$, so use options. Test option b: roots $\pm i\sqrt{6},\pm\sqrt{2}$
Product of roots: $(i\sqrt{6})(-i\sqrt{6})(\sqrt{2})(-\sqrt{2})=(6)(2)=12$, but product of roots for $x^4+2x-48=0$ is $-48$ (from Vieta: $\frac{constant}{leading coefficient}=-48$). 12≠-48, wrong.
Option a: $(2i\sqrt{2})(-2i\sqrt{2})(\sqrt{6})(-\sqrt{6})=(8)(6)=48≠-48$, wrong.
Option c: $(2i\sqrt{2})(-2i\sqrt{2})(3)(-3)=(8)(9)=72≠-48$, wrong.
Option d: $(i\sqrt{3})(-i\sqrt{3})(2\sqrt{2})(-2\sqrt{2})=(3)(8)=24≠-48$, wrong.
Wait, Vieta: product of roots is $(-1)^4*(-48)=-48$. So product of 4 roots is -48.
Option a: $(2i\sqrt{2})(-2i\sqrt{2})(\sqrt{6})(-\sqrt{6})=(8)(6)=48=-(-48)$, no.
Option b: $(i\sqrt{6})(-i\sqrt{6})(\sqrt{2})(-\sqrt{2})=(6)(2)=12≠-48$
Option c: $(2i\sqrt{2})(-2i\sqrt{2})(3)(-3)=(8)(9)=72≠-48$
Option d: $(i\sqrt{3})(-i\sqrt{3})(2\sqrt{2})(-2\sqrt{2})=(3)(8)=24≠-48$
Wait, I made mistake: Vieta for quartic $x^4+a x^3+b x^2+c x+d=0$, product of roots is $d$. So here product is $-48$.
Option a: $(2i\sqrt{2})(-2i\sqrt{2})(\sqrt{6})(-\sqrt{6})= (8)(6)=48$, so if roots are $\pm2i\sqrt{2},\pm i\sqrt{6}$, product is $(8)(6)=48$, no.
Wait, test $x=2$: $16+4-48=-28≠0$; $x=-2$:16-4-48=-36≠0; $x=\sqrt{6}$:36+2√6-48=-12+2√6≠0; $x=-\sqrt{6}$:36-2√6-48=-12-2√6≠0; $x=2i\sqrt{2}$: (2i√2)^4 +2*(2i√2)-48= 16*(4) +4i√2-48=64-48+4i√2=16+4i√2≠0; $x=i\sqrt{6}$: (i√6)^4 +2*(i√6)-48=36+2i√6-48=-12+2i√6≠0; $x=3$:81+6-48=39≠0; $x=i\sqrt{3}$: (i√3)^4 +2*(i√3)-48=9+2i√3-48=-39+2i√3≠0; $x=2\sqrt{2}$: (2√2)^4 +2*(2√2)-48=64+4√2-48=16+4√2≠0.
Wait, maybe the equation is $x^4 +2x^2 -48=0$? Then $y=x^2$, $y^2+2y-48=0$, $y=\frac{-2\pm\sqrt{4+192}}{2}=\frac{-2\pm14}{2}$, $y=6$ or $y=-8$. So $x=\pm\sqrt{6},\pm2i\sqrt{2}$, which is option a. Likely typo, assume equation is $x^4+2x^2-48=0$.

## Step2: Solve corrected quadratic in $y=x^2$
Let $y=x^2$, equation becomes:
$y^2 + 2y - 48 = 0$
Use quadratic formula:
$y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-48)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 192}}{2} = \frac{-2 \pm 14}{2}$
## Step3: Find $x$ from $y$
For $y=\frac{-2+14}{2}=6$:
$x^2=6 \Rightarrow x=\pm\sqrt{6}$
For $y=\frac{-2-14}{2}=-8$:
$x^2=-8 \Rightarrow x=\pm2i\sqrt{2}$
# Answer:
a. $\{\pm2i\sqrt{2},\pm\sqrt{6}\}$

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### Question 18 Solution
# Explanation:
## Step1: Relate root to factor
If $r$ is a root of $f(x)$, then $(x-r)$ is a factor.
## Step2: Match root to option
Given root $-4$, the factor is $(x - (-4)) = x+4$.
# Answer:
d. $x + 4$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

First Equation Solution

Step1: Test rational root candidates

Step1: Relate root to factor

Step2: Match root to option

Answer:

Question 18 Solution