Question

6h^5+0h^4-12h^3+0h^3+0h=0

Answer 1

The value of h = 0,  ± $\sqrt{2}$

Step-by-step explanation:

We have,

$6h^5+0h^4-12h^3+0h^3+0h=0$

To find, the value of h = ?

∴ $6h^5+0h^4-12h^3+0h^3+0h=0$

⇒ $6h^5$ + (0)$h^4$ - 12$h^3$ + (0)$h^3$ + (0)h = 0

⇒ $6h^5$ + 0 - 12$h^3$ + 0 + 0 = 0

⇒ $6h^5$  - 12$h^3$ = 0

Taking 6 $h^3$ as common, we get

$6h^3(h^2-2)$ = 0

⇒6$h^3$ = 0 or,   $h^2$ - 2 = 0

⇒ 6$h^3$ = 0 ⇒ h = 0

⇒ $h^2$ = 2  

⇒ h = ± $\sqrt{2}$

Hence, the value of h = 0,  ± $\sqrt{2}$