InterviewSolution
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InterviewSolution
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Answer» Correct Answer - `a to s; b to p; c to q; d to r ` `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)` |
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