What are the roots of the polynomial equation x3-7x=6x-12?

2 answers:

7 0

This is not a polynomial equation unless one of those is squared. As it stands x=-.833. If you can tell me which is squared I can help solve the polynomial.

Ok, that is usually notated as x^3 to be clear. I'll solve it now.

x^3-13x-12=0

Then use factor theorum to solve x^3-13x-12/x+1 =0
So you get one solution of x+1=0
x=-1

Then you have x^2-x-12 now you complete the square.

Take half of the x-term coefficient and square it. Add this value to both sides. In this example we have:

The x-term coefficient = −1

The half of the x-term coefficient = −1/2

After squaring we have (−1/2)2=1/4

When we add 1/4 to both sides we have:

x2−x+1/4=12+1/4

STEP 3: Simplify right side

x2−x+1/4=49/4

STEP 4: Write the perfect square on the left.

(x−1/2)2=49/4

STEP 5: Take the square root of both sides.

x−1/2=±√49/4

STEP 6: Solve for x.

x=1/2±√49/4

that is,

x1=−3

x2=4

and the one from before

x=-1

MatroZZZ [7]4 years ago

6 0

This is not a polynomial equation unless one of those is squared.

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I don't know how to do this or what i'm doing plz help

mote1985 [20]

recalling that d = rt, distance = rate * time.

we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

$\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}$

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}$

$\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}$