roots q

• 9 Feb 2018, 7:04 PM
sssona09
roots q
so I have the equation 2x^3 +3x^2 +x -5=0

when I sub in x=1/square root (y) then the answer should be 25y^3 -31y^2 +5y -4=0 but I cant seem to get that

also when I sub in x=root y, I should get 4y^3 -5y^2 +31y -25=0 but how

thank u
• 9 Feb 2018, 7:37 PM
fan96
Re: roots q
If you try to manipulate $y$ using index laws, this becomes easier.

$2\left( y^{-\frac{1}{2}}\right)^3 + 3\left(y^{-\frac{1}{2}}\right)^2 + \left(y^{-\frac{1}{2}}\right) - 5=0$

$2 y^{-\frac{3}{2}} + 3y^{-1} + y^{-\frac{1}{2}} - 5=0$

Multiply by $y^2$ to remove negative indices:

$2 y^{\frac{1}{2}} + 3y + y^{\frac{3}{2}} - 5y^2 = 0$

Shuffle terms around:

$2 y^{\frac{1}{2}} + y^{\frac{3}{2}} = 5y^2-3y$

Square both sides to remove fractional indices:

$4 y + 4y^2 + y^3 = 25y^4 - 30y^3 + 9y^2$

Clearly $y \neq 0$, so divide by $y$:

$4 + 4y + y^2 = 25y^3 - 30y^2 + 9y$

Shuffle terms again:

$25y^3 - 31y^2 + 5y-4 =0$
• 9 Feb 2018, 8:31 PM
sssona09
Re: roots q
thank you so much!!! :)