2(t^2/5) + 7t^1/5 + 3 = 0 #solve the equation by using substitution

Question

2(t^2/5) + 7t^1/5 + 3 = 0 #solve the equation by using substitution

Answer 1

2t^2/5 + 7t^1/5 + 3 = 0

2(t^1/5)² + 7t^1/5 + 3 = 0

Let u = t^1/5

2u² + 7u + 3 = 0

(2u + 1)(u + 3 = 0

u = -3

t^1/5 = -3

t = -243

u = - 1/2

t^1/5 = -1/2

t = -1/32

 

{-3, -1/32} 

 

Answer 2

Hi Lauren,

Let u = t^1/5. Then, we can rewrite the equation as follows

2u^2 + 7u + 3 = 0

(2u + 1)(u + 3) = 0

2u + 1 = 0  or u + 3 = 0

2u = -1           u = -3

u = -1/2

Let's replace u with the original expression

u = t^1/5

t^1/5 = -1/2  or t^1/5 = -3

t = -0.03125   or t = -243

The solution set is t={-0.03125, -243}