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If a, b and c are roots of the equation x^3+qx^2+rx +s=0 then the value of r is |
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Answer» 184 a+b+c=25 2a=b+2 `c^(2)=18b` Elminating a from (1) and (2), we have `b=16-(2c)/3` Then from (3), `c^(2)=18(16-(2c)/3)` or `c^(2)+12c-18xx16=0` or `(c-12)(c+24)=0` Now, c=-24 is not possible since it does not lie between 2 and 18. Hence, c=12. Then from (3), b=8 and finally from (2),a = 5. Thus, a=5,b=8 and c=12. Hence, abc`=5xx8xx12=480`. ALSO,equation `ax^(2)+bx+c=0 is 5x^(2)+8x+12=0`, which has imaginary ROOTS. If a,b,c are roots of the equation `x^(3)+QX^(2)+rx+s=0`, then the sum of PRODUCT of roots taken two at a time is `r=5xx8+5xx12+8xx12=196`. |
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