Find the minimum value of a^2+b^2
given that x^4+ax^3+bx^2+ax+1=0 has only real roots...
(The question is not very very difficult question.. if you recall the trick often used in solving 4th degree polynomials...)
62no.. i think it is far from the answer...
what was your logic btw?
1sir i just use the expansion of (x+y)4
(x+y)^{4}=^{4}C_{0}x^{4} + ^{4}C_{1}x^{3}y + ^{4}C_{2}x^{2}y^{2} +^{4}C_{3}x^{1}y^{3} + ^{4}C_{4}y^{4}
now by comparing y=1 and a=4 and b=6.
using A.M \geq G.M
\frac{{a^{2}+b^{2}}}{2}\geq \sqrt{a^{2}b^{2}}
{a^{2}+b^{2}}\geq 2ab
so answer is 48 (sorry for calculation mistake)
62but why should a=4 and b=6??
and why should this expansion be the only one possible.. ???
Refer my hint!
11Since x = 0 is not the solution of the given eqn,
Dividing by x2 in both sides , we get
x2 + ax + b + (a/x) + (1/x2) = 0
(x2+ 1/x2) + a ( x + 1/x) +b = 0 .............(1)
Putting x + 1/x = y in (1), we get
(y2 - 2) + a y + b = 0
y2 + ay + (b-2) = 0
Since it has real roots , therefore,
a 2 - 4(b-2) > 0
therefore, a 2 > 4(b-2)
Adding both sides b2, we get
a 2 + b 2 > 4 (b-2) + b2
62great work tushar.. but you still have not used the condition on y
there is some information on y that you have left out!
1sir whats the trick often used in solving 4th degree polynomials....????
62divide by x^2 ;)
there is another method of using cyclic functions of sums.. (which i read very recently!) and will put up as either a problem or some QOD soon :)
