11) can be factorised as (x-2)3 + 3(x-2)2 - 3=0
=> (a-2)(b-2)(c-2) = 3.
1)Let a,b,c be roots of the equation x3-3x2+1=0.Find (a-2)(b-2)(c-2)?
2)Suppose f(x)=x3+px2+qx+72 is divisible by both x2+ax+b and x2+bx+a(where p,q,a,b are constants and a≠b).Find the sum of squares of the roots of the cubic polynomial?
3)Find the product of uncommon real roots of the polynomials P(x)=x4+2x3-8x2-6x+15=0 and Q(x)=x3+4x2-x-10=0?
4)Let g(x)=x4+2x3-10x2+4x-10
Find g(4+3√32+√3) ?
in the last question i alrdy knw that by rationalising 4+3√32+√3 bcums 2√3 -1 and on putting it in the equation we can get the answer.....but dat is a very lengthy method.....any1 having a better solution?
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12 Answers
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211. Let us find the equation with roots α=(a-2),β=(b-2) and γ=(c-2)
So, a=α+2 so put x=x'+2 in the equation x' being a variable
(x'+2)3-3(x'+2)2+1=0
→(x'3+6x'2+12x'+8)-(3x'2+12x'+12)+1=0
→x'3+3x'2-3=0, is the required equation.
product of roots in this equation=3
so, (a-2)(b-2)(c-2)=3
2622) common root between the quadratics is 1. the other roots will come out to be 'a' and 'b'.
using the given cubic, a+b=-1 and ab=-72
hence roots are 1,8.-9 .
sum of squares = 146
1708Yes Aditiya you are Right (actually i have edited it )
for common Real Roots = -5
1057@man111 it is not easy to factorise a 4th degree quadratic....how do u do it?(hit and trial??)
71It's just heuristic. However, It can be hinted by the fact that they have some common roots. So there must be a common factor too.
@ketan,
a+b=-1 because see the quadratic with root b (or root a). There sum of roots = -a => b + 1 = -a
which is a + b = -1
341@arnab: for 1 the direct method is - the given expression = -f(2) = 3