Question

Help me plox!
\[(2+\frac{ 3 }{ x })^2-(2+\frac{3 }{ x })-12=0\]

Answer 1

do you know quadratic formula?

Answer 2

yea @study100

Answer 3

\[x= \frac{ -b \pm \sqrt b^2-4ac}{ 2a }\]

Answer 4

Yes, first, distribute everything out
\[4+ \frac{ 12 }{ x} + \frac{ 9 }{ x ^{2} } + -2 - \frac{ 3 }{ x} -12 = 0\]

Answer 5

Now you want to get rid of the fractions so multiply everything by x^2
\[x ^{2}(4+ \frac{ 12 }{ x} + \frac{ 9 }{ x ^{2} } + -2 - \frac{ 3 }{ x} -12 = 0)\]

Answer 6

\(\bf \large \begin{array}{cccllll}
a^2&-a&-12\\
&\uparrow &\uparrow \\
&-4+3&-4\cdot 3
\end{array}\)

Answer 7

Which gives you :
\[4^{2} +12 x + 9 - 2x ^{2} -3x -12x ^{2}\]

Answer 8

Combine like terms

Answer 9

= -10 x^2 + 9x + 9
\[= - (10 x ^{2} - 9x - 9)\]
a = 10
b= -9
c=- 9

Answer 10

sorry I meant
\[4 x ^{2} + 12x + 9 - 2x^{2} -3x -12x^{2}\]

Answer 11

oh ok

Answer 12

I overtyped that x under 4x^2

Answer 13

yea i got you so its -10x^2+9x+9

Answer 14

Yes, now you can use the quadratic formula, or take out negative sign
- (10x^2 -9x -9)
either way, same answer

Answer 15

then i just plug into quadratic formula

Answer 16

yea

Answer 17

Yes :)

Answer 18

x= 3/2, -3/5

Answer 19

Yes :)