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Explanation

Since, `alpha_(1), alpha_(2) " are the roots of "ax^(2) + bx +c =0`
`rArr alpha_(1) + alpha_(2) = -(b)/(a) " and " alpha_(1)alpha_(2) = (c)/(a)" "...(i)`
Also, `beta_(1), beta_(2) " are the roots of "px^(2) + qx + r =0`
`rArr beta_(1) + beta_(2) = -(q)/(p) " and" beta_(1)beta_(2) = (r)/(p) " "...(ii)`
Given system of equtions
`alpha_(1)y + alpha_(2)z =0`
and `beta_(1) y + beta_(2) z = 0`, has non-trivial solution.
`therefore |{:(alpha_(1), alpha_(2)), (beta_(1), beta_(2)):}| = 0rArr (alpha_(1))/(alpha_(2)) = (beta_(1))/(beta_(2))`
Applying compenendo-dividendo, `(alpha_(1) + alpha_(2))/(alpha_(1) -alpha_(2)) = (beta_(1) +beta_(2))/(beta_(1) - beta_(2))`
`rArr (alpha_(1) + alpha_(2)) (beta_(1)-beta_(2)) = (alpha_(1) -alpha_(2))(beta_(1) + beta_(2))`
`rArr (alpha_(1) + alpha_(2))^(2) {(beta_(1) +beta_(2))^(2) -4beta_(2)beta_(2)}`
`=(beta_(1) + beta_(2))^(2){(alpha_(1) + alpha_(2))^(2) -4alpha_(1), alpha_(2)}`
From Eqs. (i) and (ii), we get
`(b^(2))/(a^(2)) ((q^(2))/(p^(2)) - (4r)/(p)) = (q^(2))/(p^(2))((b^(2))/(a^(2)) - (4c)/(a))`
`rArr (b^(2)q^(2))/(a^(2)p^(2)) -(4b^(2)r)/(a^(2)p) = (b^(2)q^(2))/(a^(2)p^(2)) - (4q^(2)c)/(ap^(2))`
`rArr (b^(2)r)/(a) = (q^(2)c)/(p) rArr (b^(2))/(q^(2)) = (ac)/(pr)`