Sup Veeky Forums i have this really basic question in my exam guide, it should be really easy but o think the teacher wrote it wrong but maybe im just wrong so try to solve it guys ill post the answer in 15 minutes pld post the procedure btw
Sup Veeky Forums i have this really basic question in my exam guide...
[eqn]
3 \left( 24x^{2} + 26x - 15 \right)
[/eqn]
I arrived there too and got it wrong btw you have to solve for x
...
ok I think you guys confirmed the exercise is wrong, the final answer is supposed to be 78
Second part
>solve for x
You only posted a shitty snapshot, so I typed it in wolfram alpha and pasted whatever it spat out.
ok I thought it was obvious but try it, hopefully I am wrong
solutions are: -1.5 and 0.4166667
1) Simplify everything up to : 72x^2+78x-45
2) Factor out the quadratics
3) (6x+9)(12x-5)=0
6x+9=0 , 12x-5=0
6x=-9 , 12x = 5
x=-9/6 , x=5/12
4) Simplify answer to : -3/2 and 5/12
Winrar
Ok try to simplify this now, sorry for the shitty quality
(x+4)(x-1) / (x+1)(x-1)
oops I mistyped
[math](x+4)(x+1)/(x+1)(1-x)[/math]
[eqn]
\begin{aligned}
3 \left( 24x^2 + 26x - 15 \right) &= 0 \\
3 \left( (2x+3)(12x+5) \right) &= 0 \\
x &= \left \{ -\frac{3}{2} , -\frac{5}{12} \right \}
\end{aligned}
[/eqn]
pls fug off now
[math] \frac{-4x^{4}+13x^{2}-9}{2x^{3}+3x^{2}-17x+12} [/math]
Did in a hurry, not sure.
I may be new to this board, but I could understand the sticky.
[eqn]
\frac{-(2x+3)(x+1)}{x+4}
[/eqn]
wrong
right pls post your procedure
I was getting lazy with LaTeX resizing and align (hence why I only posted the solution) but whatever I'll show the work.
[eqn]
\begin{align}
\frac{4x^2 - 9}{x^2 + 3x - 4} \cdot \frac{1 - x^2 }{2x - 3} &= \frac{(2x+3)(2x-3)}{(x-1)(x+4)} \cdot \frac{(1+x)(1-x)}{2x - 3} \\
&= \frac{(2x+3)(2x-3)}{(x-1)(x+4)} \cdot \frac{-(x+1)(x-1)}{2x - 3} \\
&= \frac{-(2x+3)(x+1)}{x+4}
\end{align}
[/eqn]
I would literally suck your cock thanks senpai
why did you add a - before (x+1)(x-1)
he factored out a negative from (1-x)
[math](1-x^2) = (1-x)(1+x) = -(-1+x)(1+x) = -(x-1)(x+1)[/math]
[math](1-x)^2[/math] should be the first one, sorry
You can't just add =0 to the end
What said. If I didn't factor out the negative, I wouldn't be able to cancel as much terms. Just remember to try that out whenever you have stuff like [math](a+x)(a - x)[/math].