Question

2x ^4 +15x 2 −50=0

Answer 1

put a t instead of X^2----> 2t^2+15t-50=0
then u know..as usually with D=b^2-4ac

Answer 2

the answer in the book says \[\pm \sqrt{10}\] then a 2 underneath it

Answer 3

D=625>0 so we have 2 solutions 4 t
X1=-b

Answer 4

t1=(-b+sqrD)/2a=(-15+25)/4=-10/4
t2=(-b-sqrD)/2a=(-15-25)2a=-35/4
then ....t=X^2---> x=sqrt

Answer 5

this is wht i think

Answer 6

Thanks!

Answer 7

t1=(-b+sqrD)/2a=(-15+25)/4=10/4*********
oh w8...sqr of a negative no doesn't exist so it's only t1 a solution 4 this

Answer 8

Can you help me with this one \[\sqrt[3]{25x}{5}^y^{2}\]

Answer 9

write it again plsss

Answer 10

\[\sqrt[3]{25x ^{5}y ^{2}}\]

Answer 11

wht should i do here?

Answer 12

oh I forgot the rest of the problem! next to that one is \[\sqrt[3]{10x ^{6}y ^{8}}\]

Answer 13

The answer in the book is \[5x ^{3}y ^{3} \sqrt[3]{2x ^{2}y}\]

Answer 14

not sure how they got it...