1.

If a, b and c are roots of the equation x^3+qx^2+rx +s=0 then the value of r is

Answer»

184
196
224
none of these

Solution :We have,
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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