Saved Bookmarks
| 1. |
All possible roots of polynomial with integral coefficients can be identified by "RATIONAL ROOT TEST" according to rational root test if a plynomial ltbr. `a_(n)p^(n)+a_(n-1)p^(n-1)q+a_(n-2)p^(n-2)q^(2)+`…..`+a_(2)p^(2)q^(n-2)+a_(1)pq^(n-1)+a_(0)q^(n)=0` Every term in above equation except possible the last one is divisible by `p` hence `p` should also divide `a_(0)q^(n)`, since `q` and `p` are relatively prime `p` must divide `a_(0)`. Similarly `q` also divides `a_(n)`. Now consider an equation `6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0` and answer the following questions Q. if given equation and equation `x^(3)+ax^(2)+ax+1=0` have two roots common then possible value(s) of is/are?A. `0`B. `-2`C. `-3`D. `3` |
|
Answer» Correct Answer - D (13 to 14) `6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0` rational roots of equation `x=+-1,+-(1)/(2),+-(1)/(2),+-(1)/(6)` are possible `x=(1)/(6)` satisfies the equation `6x^(5)-x^(4)-18x^(4)+3x^(3)-12x^(3)+2x^(2)-18x^(2)+3x+6x-1=0` `(6x-1)(x^(4)-3x^(3)-2x^(2)-3x+1)=0` `x^(4)-3x^(3)-2x^(2)-3x+1)=0` `x^(4)-3x^(3)-2x^(2)-3x+1=0` `x^(2)+(1)/(x^(2))-3(x+(1)/(x))-2=0` `(x+(1)/(x))^(2)-3(x+(1)/(x))-4=0` `t^(2)-3t-4=0` `(t-4)(t+1)=0` `x+(1)/(x)=4` or `x+(1)/(x)+1=0` `x^(2)-4x+1=0` `x^(2)+x+1=0` sum of real roots will be `4+(1)/(6)=(25)/(6)` for rational values of a roots of cubic `x^(2)+ax^(2)+ax+1=0` will be in conjugate pair so either `x^(2)-4x+1=0` or `x^(2)+x+1=0` is a factor of `x^(3)+ax^(2)+1=0` if `x^(2)-4x+1=0` is a factor of `x^(3)+ax^(2)+ax+1=0` then `a=-3` and `x^(2)+x+1=0` is a factor of `x^(3)+ax^(2)+ax+1=0` then `a=2` |
|