Question

how do you solve x^2 - 4x?

Answer 1

what means "solve" in this case? you do not have an equal sign

Answer 2

solve?

Answer 3

its # 4 im stuck on

Answer 4

oh x^2-4x=0

Answer 5

we can solve x^2-4x=0

Answer 6

x(x-4)=0
x=0, x=4

Answer 7

....... I feel stupid.

Answer 8

oh you aren't stupid

Answer 9

how about # 6?

Answer 10

you corrected me earlier on cheat cheat lol

Answer 11

OH Your the same person?! im new to this website!

Answer 12

\[x^3-x^2-x=0\]
\[x(x^2-x-1)=0\]
\[x=0, x^2-x-1=0\]

Answer 13

just use the quadratic formula for the second equation

Answer 14

but it dont work

Answer 15

why not?

Answer 16

\[x=\frac{1 \pm \sqrt{1-4(1)(-1)}}{2(1)}=\frac{1 \pm \sqrt{5}}{2}\]

Answer 17

so you have three real solutions!

Answer 18

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=1,b=-1,c=-1\]

Answer 19

OH! i thought it had to be a whole #

Answer 20

can u help me with # 10

Answer 21

\[\frac{\frac{1}{x+1}}{x}=20\]
\[\frac{1}{x(x+1)}=20\]
\[1=20x(x+1)\]
\[20(x^2+x)=1\]
\[20x^2+20x-1=0\]

Answer 22

put that into the quad. equation right?

Answer 23

\[x=\frac{-20 \pm \sqrt{20^2-4(20)(-1)}}{20}=\frac{-20 \pm \sqrt{400+80}}{20}\]
\[=\frac{-20 \pm \sqrt{480}}{20}=\frac{-20 \pm \sqrt{16 \cdot 30}}{20}=\frac{-20 \pm 4\sqrt{30}}{20}\]
\[=\frac{4(-5 \pm \sqrt{30})}{4 \cdot 5}=\frac{5 \pm \sqrt{30}}{5}\]

Answer 24

yes

Answer 25

And # 14 kinda stumped me.

Answer 26

\[2\sqrt{x}=x-3\]
square both sides
\[(2\sqrt{x})^2=(x-3)^2\]
\[4x=x^2-6x+9\]
\[x^2-6x-4x+9=0\]
\[x^2-10x+9=0\]
\[x^2-x-9x+9=0\]
\[x(x-1)-9(x-1)=0\]
\[(x-1)(x-9)=0=>x=1 or x=9\]
----
we must check these because one of them or both may not be a solution
\[2\sqrt{1} \neq 1-3\]
so x=1 is not a solution
\[2\sqrt{9}=9-3\]
so x=9 is a solution

Answer 27

what about #15? that looked hard cause the cube root sign to me

Answer 28

oops i already exited out of the attachment let me see if i can find it again

Answer 29

\[\sqrt[3]{2x+1}-4=-1\]
\[\sqrt[3]{2x+1}=4-1\]
\[\sqrt[3]{2x+1}=3\]
to gt rid of cube root cube both sides
\[2x+1=3^3\]
\[2x=27-1\]
\[x=\frac{26}{2}=13\]

Answer 30

and # 11?

Answer 31

\[\frac{x}{x+2}-\frac{2}{2x-1}=\frac{1}{5}\]
get rid of the fractions
\[\frac{x}{x+2}5(x+2)(2x-1)-\frac{2}{2x+1}5(x+2)(2x-1)=\frac{1}{5}5(x+2)(2x-1)\]
\[5x(2x-1)-10(x+2)=(x+2)(2x-1)\]
\[10x^2-5x-10x-20=2x^2-x+4x-2\]
\[10x^2-2x^2-5x+x-4x-10x-20+2=0\]
\[8x^2-18x-18=0\]
\[4x^2-9x-9=0\]
use the quad formula

Answer 32

# 15 seems so easy now. I feel dumb.

Answer 33

i'm gonna take a break
you should make a new thread if you have more questions

Answer 34

good lord looka that latex!

Answer 35

one more then i will stop. Please.

Answer 36

you hit the question jackpot here myin

Answer 37

and the next one is # 16.

Answer 38

just post it. you will get lots of answers

Answer 39

\[1+\frac{-2x}{\sqrt{25-x^2}}=0\]
\[\sqrt{25-x^2}+(-2x)=0\] <--- i multiplied both sides by the sqrt(25-x^2}
\[\sqrt{25-x^2}=2x\]
square both sides
\[25-x^2=(2x)^2\]
\[25-x^2=4x^2\]
\[4x^2+x^2-25=0\]
\[5x^2-25=0\]
\[x^2-5=0\]
\[x^2=5=>x=\pm \sqrt{5}\]

Answer 40

can i go now
i need a break :(

Answer 41

THANK YOU VERY MUCH. and yes you can leave. :D

Answer 42

thanks for your permission lol

Answer 43

LOL