Question

`15x^2-28=x` 

 

Answer 1

Given: 

`15x^2-28=x` 

⇒`15x^2-x-28=0` 

On comparing it with `ax^2+bx+c=0` we get; 

`a=25,b=-1  and c=-28` 

Discriminant D is given by: 

`D=(b^2-4ac)` 

=`(-1)^2-4xx15xx(-28)` 

=`1-(-1680)` 

=`1+1680` 

=`1680` 

=`1681>0` 

Hence, the roots of the equation are real.
Roots α and β are given by: 

`α=(-b+sqrt(D))/(2a)=(-(-1)+sqrt(1681))/(2xx25)=(1+41)/30=42/30=7/5` 

 `β=(-b-sqrt(D))/(2a)=(-(-1)-sqrt(1681))/(2xx25)=(1-41)/30=-40/30=(-4)/3` 

Thus, the roots of the equation are `7/5` and `-4/3`