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It is given that the polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Let Q(x) = x2- 2x 22. It is also given that P(Q(x)) has no real roots then a is equal to (A) -45 (B)-55 (C) 45 (D) 60?
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It is given that the polynomial P(x) = x3 ax2 bx c has three dis...
Given Information:

Polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Q(x) = x2- 2x 22. P(Q(x)) has no real roots.


To Find:

Value of a in P(x).


Solution:

Let the roots of P(x) be p, q, and r. Then we have:


  • p + q + r = a

  • pqr = -c

  • p + q + r = a

  • pq + qr + rp = b



Since P(x) has three distinct positive integer roots, we can assume that p < q="" />< r="" are="" positive="" integers.="" we="" are="" also="" given="" that="" p(22)="21," which="" means="" />


P(22) = (22 - p)(22 - q)(22 - r) = 21


Since 22 - p, 22 - q, and 22 - r are positive integers, we can only have:


22 - p = 1, 3, or 7


22 - q = 1, 3, or 7


22 - r = 1, 3, or 7


Without loss of generality, we can assume that:


22 - p = 1, 22 - q = 3, and 22 - r = 7


which gives us:


p = 21, q = 19, and r = 15


Now, let us consider P(Q(x)):


P(Q(x)) = (Q(x) - p)(Q(x) - q)(Q(x) - r)


Substituting Q(x) = x2 - 2x + 22, we get:


P(Q(x)) = (x2 - 2x + 22 - p)(x2 - 2x + 22 - q)(x2 - 2x + 22 - r)


Expanding the above expression and simplifying, we get:


P(Q(x)) = x6 - (p + q + r + 6)x4 + (pq + qr + rp + 8(p + q + r) + 44)x2 - pqr - 44(p + q + r) - 484


Substituting the values of p, q, and r, we get:


P(Q(x)) = x6 - 72x4 + 1330x2 - 13515


Since P(Q(x)) has no real roots, its discriminant is negative:


72^2 - 4(1330) < />


which gives us:


a = p + q + r = 21 + 19 + 15 = 55


Answer:
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It is given that the polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Let Q(x) = x2- 2x 22. It is also given that P(Q(x)) has no real roots then a is equal to (A) -45 (B)-55 (C) 45 (D) 60?
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It is given that the polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Let Q(x) = x2- 2x 22. It is also given that P(Q(x)) has no real roots then a is equal to (A) -45 (B)-55 (C) 45 (D) 60? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about It is given that the polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Let Q(x) = x2- 2x 22. It is also given that P(Q(x)) has no real roots then a is equal to (A) -45 (B)-55 (C) 45 (D) 60? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is given that the polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Let Q(x) = x2- 2x 22. It is also given that P(Q(x)) has no real roots then a is equal to (A) -45 (B)-55 (C) 45 (D) 60?.
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