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Exercise 2.4 question number 3rd |
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Answer» Comparing given polynomials with px3 + qx2 + rx + t .So, p=1 , q= -3 , r=1 & t=1 Sum of zeros = a-b , a , a+b -q/p =3a -(-3)/1 = 3a 3 = 3a a = 1 The zeros are 1-b,1,1+b Multiple of zeros are = 1(1-b)(1+b) -t/p = 1-b2 -1/1 = 1-b2 1-b2 =-1 1+1 = b2 b2 =2 b = √2 b = +- √2 So, the zeros are a=1 & b=√2 & b=-√2 the polynomial\xa0.\xa0=\xa0\xa0And\xa0=\xa0\xa0\xa0\xa0\xa0\xa0  \xa0Hence\xa0\xa0and\xa0. |
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