(x^2-x-9)^3/4-13=14
solve the equation with rational exponents

Question

myininaya · Answer

$$(x^2-x-9)^\frac{3}{4}=27$$
$$(x^2-x-9)=27^\frac{4}{3}$$
$$(x^2-x-9)=3^4$$
$$x^2-x-9=81$$
$$x^2-x-90=0$$
$$(x-10)(x+9)=0$$

myininaya · Answer

$$x=10 or x=-9$$

anonymous · Answer

add 13 to both sides is a start. get 
$$(x^2-x-9)^{\frac{3}{4}}=27$$
take the cube root get 
$$\sqrt[4]{(x^2-x-9)}=3$$ raise to the power of 4 get 
$$x^2-x-9=81$$ then solve

anonymous · Answer

or be quick like myininaya (show off!)

myininaya · Answer

check both since you have an even root

myininaya · Answer

obviously 10 is god
now check -9

myininaya · Answer

and -9 is good

anonymous · Answer

aren't you tired after a long day of calc IX?

myininaya · Answer

i'm suppose to be reading something

anonymous · Answer

torturing your poor dears by making the find derivatives using triangles?  you cover series in that class?

myininaya · Answer

no
series is cover in our cal 3 class
i need to review that stuff
its been forever since i talked about convergence of a series

anonymous · Answer

too bad they took the one fun thing out and left those stultifyingly boring "techniques of integration" in 
entire content of class on the back cover of stewart

myininaya · Answer

oh yeah i had a nap

anonymous · Answer

zzzzzzzzzzzzzzz