Question

Help needed!! Don’t know how to do!!

Answer 1

Answer:

D

Step-by-step explanation:

1) Factorise x

x(2x^4-3x^2-20) = 0

a product is equal to 0 when at least one factor is equal to 0.

so a solution is

x = 0

we have now to solve

2x^4 - 3x^2 - 20 = 0

we can put

x^2 = y

and we have

2y^2 - 3y - 20 = 0

it is now an equation of second degree

Δ = 9 + 160 = 169

y1 = (3 + 13)/4 = 4

y2 = (3 - 13)/4 = -10/4 = -5/2

for the previous relationship we have

x^2 = 4

x = 2

x = -2

x^2 = -5/2

a square number can’t be negative, so this equation is impossible.

In conclusion we can say that the original equation has three solution:

0, -2, 2

Answer 2

Answer:

D,0,2,-2

Step-by-step explanation:

2x^5-3x^3-20x=0

x(2x^4-3x^2-20)=0

x=0

or 2x^4-3x^2-20=0

put x²=t

2t²-3t-20=0

-20×2=-40

8-5=3

8×-5=-40

2t²-(8-5)t-20=0

2t²-8t+5t-20=0

2t(t-4)+5(t-4)=0

(t-4)(2t+5)=0

t=4

x²=4

x=2,-2

t=-5/2

x²=-5/2

it gives imaginary  root. so real rational roots are 0,2,-2