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.!