Question

How to solve any of these equations in comments?

Answer 1

\[\frac{ 7x^2 +5x}{x^2 +1}-\frac{ 5x }{x^2-6 }=0\]

Answer 2

\[\frac{ x-1 }{(x+1)(3x-5)\sqrt{x+3} }=0\]

Answer 3

\[(x+1)^\frac{ -1 }{2} +\frac{ (x-1)(x+1)^\frac{ 1 }{ 2 } }{ \sqrt{x^3 +1} }=0\]

Answer 4

\(\bf \cfrac{ 7x^2 +5x}{x^2 +1}-\cfrac{ 5x }{x^2-6 }=0\implies \cfrac{ 7x^2 +5x}{x^2 +1}=\cfrac{ 5x }{x^2-6 }
\\ \quad \\
(x^2-6)(7x^2 +5x)=(x^2 +1)(5x)\)

once you simplify that, you'd left with a ternary

have you covered the "rational root test" yet?

Answer 5

Ah, that makes sense. No we haven't covered it yet.

Answer 6

well... you'd need it to solve the ternary , so maybe you're not required to get "x" but just to simplify it

Answer 7

well the problems specifically say solve for x... this is actually summer work for ap calculus. we may have gone over the rational root theorem earlier last year but I don't fully remember it. It does sound familiar though.

Answer 8

hmm this is for calculus? hmm

Answer 9

ok... lemme do a quick rundown over it

Answer 10

yup. ok. thanks for the help.

Answer 11

\(\bf \cfrac{ 7x^2 +5x}{x^2 +1}-\cfrac{ 5x }{x^2-6 }=0\implies \cfrac{ 7x^2 +5x}{x^2 +1}=\cfrac{ 5x }{x^2-6 }
\\ \quad \\
(x^2-6)(7x^2 +5x)=(x^2 +1)(5x)
\\ \quad \\
7x^4\cancel{ +5x^3 }-42x^2-30x=\cancel{ 5x^3 }+5x\implies 7x^4-42x^2-35x=0
\\ \quad \\
7x(x^3-6x-5x)=0\implies {\color{brown}{ 1}}x^3-6x-{\color{blue}{ 5}}x=0
\\ \quad \\
\textit{rational root test, likely roots are }\pm \cfrac{{\color{blue}{ 5,1}}}{{\color{brown}{ 1}}}
\\ \quad \\
\textit{so checking by say using synthetic division, using }-1
\\ \quad \\
x^3+0x^2-6x-5x \div x+1\qquad
\begin{array}{ccccc}
-1& 1& 0&-6&-5\\
&&-1&+1&+5
\\\hline\\
&1&-1&-5&0
\end{array}\)

so as you can see, the -1, that is x+1=0 => x=-1

give us a remainder of 0 in the synthetic division

so we can say that -1 is root

Answer 12

so our quotient of 1, -1, -5 will turn into \(\bf x^2-x-5\)

meaning that \(\bf x^3+0x^2-6x-5x \div x+1\implies x^2-x-5\qquad thus
\\ \quad \\
x^3-6x-5x=0\iff (x^2-x-5)(x+1)=0\)

Answer 13

ah i see, I was actually thinking about synthetic but It seemed daunting to attempt to try many numbers. the rational root theorem will help alot.

Answer 14

so the quadratic factor is simple to break down, by just using the quadratic formula

Answer 15

the 2nd one is ... almost a given \(\bf \cfrac{ x-1 }{(x+1)(3x-5)\sqrt{x+3} }=0\implies x-1=0[(x+1)(3x-5)\sqrt{x+3}]
\\ \quad \\
x-1=0\implies x=1\)

Answer 16

ok when i enter the second one into mathway.com it will give me \[x \sqrt{x+3}-\sqrt{x+3}=0 \] even though i tell it to solve for x.

Answer 17

But to me that doesn't make sense since if i were going to solve I would just as you had an thus get the same answer as you.

Answer 18

right

Answer 19

maybe ... posting the 2nd and 3rd anew, someone else may know

I checked in http://www.wolframalpha.com/input/?i=%28x%2B1%29%2F%28%28x%2B1%29%283x-5%29sqrt%28x%2B3%29%29%3D0
and it gave me "no solution"

then I checked in my graphic calculator... it has 1 vertical and 1 horizontal asymptotes
never touching the x-axis.. meaning no solution

Answer 20

Ok. The second one has stumped me for a few days cuz it seems so simple yet that seems too easy. Thanks for the help though.