Q.
Let a,b and c be three distinct real roots of the cubic x3+2x2−4x−4=0.
If the equation x3+qx2+rx+s=0 has roots a1,b1 and c1, then the value of (q+r+s) is equal to
184
89
Complex Numbers and Quadratic Equations
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Solution:
We have x3+2x2−4x−4=0 has roots, a,b and c.....(1)
On replacing x by x1 in equation (1), we get 4x3+4x2−2x−1=0⇒x3+x2−21x−41=0,
which has roots a1,b1 and c1. ....(2) ∴ On comparing equation (2) with x3+qx2+rx+s=0, we get q=1,r=2−1,s=4−1
Hence (q+r+s)=41