## anonymous 5 years ago 2x ^4 +15x 2 −50=0

1. angela210793

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

2. anonymous

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

3. angela210793

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

4. angela210793

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

5. angela210793

this is wht i think

6. anonymous

Thanks!

7. angela210793

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

8. anonymous

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

9. angela210793

write it again plsss

10. anonymous

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

11. angela210793

wht should i do here?

12. anonymous

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

13. anonymous

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

14. anonymous

not sure how they got it...