Let `alpha_1,alpha_2` and `beta_1, beta_2` be the roots of the equation `ax^2+bx+c=0` and `px^2+qx+r=0` respectively. If the system of equations `alpha_1y+alpha_2z=0` and `beta_1y_beta_2z=0` has a non trivial solution then prove that `b^2/q^2=(ac)/(pr)`

Explanation

Clearly,
`alpha_(1)+alpha_(2) =- b//a ,alpha_(1)alpha_(2) =c//a`
`" and " beta_(1)+beta_(2) =- q//p , beta_(1)beta_(2)=r//p`
System of equations `alpha_(1)y+alpha_(2)z= 0,beta_(a)y+beta_(2)z=0` has a non-trivial solution.
`" So "|underset( beta_(1)" "beta_(2))(alpha_(1)" "alpha_(2))|=0`
`i.e., " " alpha_(1)beta_(2)-alpha_(2)beta_(1)=0`
`" or " (alpha_(1))/(alpha_(2)) =(beta_(1))/(beta_(2))`
`rArr (alpha_(1))/(beta_(1)) =(alpha_(2))/(beta_(2))=(alpha_(1)+alpha_(2))/(beta_(1)+beta_(2)) = sqrt((alpha_(1)alpha_(2))/(beta_(1)beta(2)))`
` rArr (alpha_(1)+alpha_(2))/(beta_(1)+beta_(2))=sqrt((alpha_(1)alpha_(2))/(beta_(1)beta_(2)))`
`rArr (-b//a)/(-q//p)=sqrt((c//a)/(r//p))`
`rArr b^(2) pr= q^(2)ac`