Consider polynomial ax^2+bx+c such that p^2(0)+p^2(1)+p^2(2)+166=6p(0)+12p(1)+22p(2).then find whether foloeing are correct (A) a-b+c=2.(B)min of p(x)=2?

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Answer 1 · 2018-07-16T06:52:22

Both (A) and (B) are True.

I presume, #p(x)=ax^2+bx+c, and, p^2(x)=(p(x))^2#.

Now, given that,

#p^2(o)+p^2(1)+p^2(2)+166=6p(o)+12p(1)+22p(2)#.

#:. p^2(0)-6p(0)+p^2(1)-12p(1)+p^(2)-22p(2)=-166#.

On completing squares,

#{p(0)-3}^2+{p(1)-6}^2+{p(2)-11}^2#,

#=9+36+121-166#.

#:. {p(0)-3}^2+{p(1)-6}^2+{p(2)-11}^2=0#.

#:. {p(0)-3}=0, {p(1)-6}=0, &, {p(2)-11}=0#.

#:. p(0)=3, p(1)=6, &, p(2)=11#.

Now, #p(0)=3 rArr c=3...............(star_1)#.

#p(1)=6 rArr a+b+c=6."Then, by "(star_1), a+b=3....(star_2)#.

#"Similarly, from "p(2)=11," we get, "2a+b=4...(star_3)#.

#"Altogether "(star_1), (star_2) and (star_3) rArr (a,b,c)=(1,2,3)#.

Consequently, #a-b+c=2#, showing that, (A) is true.

Finally, to find #p_min#, let us observe that,

#p(x)=x^2+2x+3=(x+1)^2+2#.

#because AA x in RR, :. (x+1)^2 ge 0, &, :., (x+1)^2+2 ge 2#.

Clearly, #p_min=2#, meaning that, (B) is also true.

Enjoy Maths.!