Question

Need help with a radical equation! Will give medal!

Answer 1

\[\sqrt{x^2+1}+\frac{ 8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}\]

Answer 2

well... taking a peek at the left-side
if we were to add those two, what would be the LCD?

Answer 3

i hope this helps http://www.mathportal.org/calculators.php

Answer 4

\(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }\implies \cfrac{\square ?+\square ?}{{\color{brown}{ \sqrt{x^2+1}}}}\)

Answer 5

LCD is the square root of x^2+1

Answer 6

yeap

Answer 7

so notice if we add those two then -> \(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }\implies \cfrac{(\sqrt{x^2+1})(\sqrt{x^2+1})+8}{{\color{brown}{ \sqrt{x^2+1}}}}
\\ \quad \\
\cfrac{ (\sqrt{x^2+1})^2+8 }{ \sqrt{x^ 2+1} }\implies \cfrac{ x^2+1+8 }{ \sqrt{x^ 2+1} }
\)

Answer 8

so you'd end up with \(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}
\\ \quad \\
\cfrac{ x^2+1+8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}\implies x^2+1+8 =(\sqrt{x^ 2+1})(\sqrt{x^2+9})
\\ \quad \\
x^2+9=\sqrt{(x^ 2+1)(x^2+9)}\)

Answer 9

Where did the other x^2+1 come from on the top right

Answer 10

well... actaully

Answer 11

hmmm which one? \(\bf \cfrac{(\sqrt{x^2+1})({\color{blue}{ \sqrt{x^2+1}}})+8}{{\color{brown}{ \sqrt{x^2+1}}}}\leftarrow {\color{blue}{ \textit{this one?}}}\)

Answer 12

yes

Answer 13

\(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }\implies \cfrac{\sqrt{x^2+1}}{1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }\implies \cfrac{\square ?+\square ?}{{\color{brown}{ \sqrt{x^2+1}}}}\)

notice the LCD
if you divide the LCD by 1...you'd get? \(\bf \sqrt{x^2+1}\) thus

and when you divide the LCD on the 2nd fraction by it \(\bf \sqrt{x^ 2+1}\) you'd get 1

Answer 14

ok please continue

Answer 15

so one could say that \(\bf \cfrac{\sqrt{x^2+1}}{1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }\implies \cfrac{(\sqrt{x^2+1})(\sqrt{x^2+1})+8(1)}{{\color{brown}{ \sqrt{x^2+1}}}}\)

Answer 16

ok.. so lemme redo the above some \(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}
\\ \quad \\
\cfrac{ x^2+1+8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}\implies x^2+9 =(\sqrt{x^ 2+1})(\sqrt{x^2+9})
\\ \quad \\
\cfrac{x^2+9}{\sqrt{x^2+9}}=x^2+1\)

see it thus far?

Answer 17

hmmm forgot the root there... leme fix that \(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}
\\ \quad \\
\cfrac{ x^2+1+8 }{ \sqrt{x^ 2+1} }=\sqrt{x^2+9}\implies x^2+9 =(\sqrt{x^ 2+1})(\sqrt{x^2+9})
\\ \quad \\
\cfrac{x^2+9}{\sqrt{x^2+9}}=\sqrt{x^2+1}\)

see it thus far?

Answer 18

yes

Answer 19

notice the left-side
top and bottom look quite the same
just different exponentials

thus \(\bf \large {\cfrac{x^2+9}{\sqrt{x^2+9}}=\sqrt{x^2+1}\qquad {\color{brown}{ x^2+9}}=a\qquad then
\\ \quad \\
\cfrac{{\color{brown}{ a}}^1}{{\color{brown}{ a}}^{\frac{1}{2}}}=\sqrt{x^2+1}\implies a^1\cdot a^{-\frac{1}{2}}=\sqrt{x^2+1}
\\ \quad \\
a^{1-\frac{1}{2}}=\sqrt{x^2+1}
}\)

see it thus far?

Answer 20

yes

Answer 21

ok so hmmm

Answer 22

got a funky value... one sec

Answer 23

ok

Answer 24

this is what I got thus far \(\bf \large {
\cfrac{x^2+9}{\sqrt{x^2+9}}=\sqrt{x^2+1}\qquad {\color{brown}{ x^2+9}}=a\qquad then
\\ \quad \\
\cfrac{{\color{brown}{ a}}^1}{{\color{brown}{ a}}^{\frac{1}{2}}}=\sqrt{x^2+1}\implies a^1\cdot a^{-\frac{1}{2}}=\sqrt{x^2+1}
\\ \quad \\
a^{1-\frac{1}{2}}=\sqrt{x^2+1}\implies a^{\frac{1}{2}}=\sqrt{x^2+1}
\\ \quad \\
{\color{brown}{\sqrt{x^2+9} }}=\sqrt{x^2+1}\implies

x^2+9=x^2+1 }\)

Answer 25

but...that sorta cancels out \(x^2\) and leave us want a non-equation of 9=1

Answer 26

which simply means as far as I can tell that \(\bf \sqrt{x^2+1}+\cfrac{ 8 }{ \sqrt{x^ 2+1} }\ne\sqrt{x^2+9}\)