Match the following lists:

Answer

`a to s; b to p; c to q; d to r `

Explanation

`atos,bto""p,ctoq,d""tor`
(a) `x^(2)ax+b=0` has root `lapha`. Hence.
`alpha^(2)+aalpha+ b=0" "(1)`
`x^(2)+px+q=0` has root `-alpha`. Hence,
`alpha^(2)-palpha+q=" "(2)`
Eliminating `alpha` from (1) and (2), we get
`(q-b)^(2)=(aq+bp)(-p-a)`
or `(q-b)^(2)=-(aq+pb)(p+a)`
(b) `x^(2)ax+b=0` has root `alpha`. Hence,
`alpha^(2)+aalpha+b=0 " "(1)`
`x^(2)+px+q=0` has root `1//alpha`. Hence,
`qalpha^(2)+palpha+1=0" "(2)`
Eliminating `alpha` from (1) and (2), we get
`(1-bq)^(2)=(a-pb)(p-aq)`
(c) `x^(2)+ax+b=0` has roots, `alpha,beta`. Hence,
`alpha^(2)+aalpha+b=0" "(1)`
`x^(2)+px+q=0` has roots `-2//alpha,gamma`. Hence,
`qalpha^(2)-2palpha+4=0" "(2)`
Eliminating `alpha` from (1) and (2), we et
`(4-bq)^(2)=(4a+2pb)(-2p-aq)`
(d) `x^(2)+ax+b=0` has roots `alpha,beta`. Hence,
`alpha^(2)+aalpha+b=0" "(1)`
`x^(2)+px+q=0` has roots `-1//2alpha,gamma`. Hence,
`4qalpha^(2)-2palpha+1=0" "(2)`
Elimintaing `alpha` from (1) and (2), we get
`(1-4bq)^(2)=(a+2bp)(-2p-4aq)`