Question

(y^2-5y)^2-20(y^2-5y)+84=0 find the solution of the quadratic equation by factorization method​

Answer 1

Given:

$(y^2-5y)^2-20(y^2-5y)+84=0$

To prove:

solve by factorization method​

Solution:

$(y^2-5y)^2-20(y^2-5y)+84=0$

let $(y^2-5y) = t$

$\Rightarrow t^2-20t+84=0\\\\\Rightarrow t^2-(14+6)t+84=0\\\\\Rightarrow t^2-14t-6t+84=0\\\\\Rightarrow t(t-14)-6(t-14)=0\\\\\Rightarrow (t-14)(t-6)=0$

put the t value:

$\Rightarrow (t-14)(t-6)=0\\\\\Rightarrow ((y^2-5y)-14)((y^2-5y)-6)=0\\\\\Rightarrow y^2-5y-14=0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ y^2-5y-6=0\\\\\Rightarrow y^2-(7-2)y-14=0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ y^2-(6-1)y-6=0\\\\\Rightarrow y^2-7y+2y-14=0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ y^2-6y+y-6=0\\\\\Rightarrow y(y-7)+2(y-7)=0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ y(y-6)+1(y-6)=0\\\\\Rightarrow (y-7)(y+2)=0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ (y-6)(y+1)=0$

$\Rightarrow y= 7, -2 \ \ \ \ \ \ \ \ and \ \ \ \ y = 6, -1$

The final value is 7,-2 and 6,-1.