Q.55. Polynomial equation.

Question 55.
The equation P(x) = x4 – 16x3 + 94x2 +px + q = 0 has two double roots. Find p+q.

Answer 55.
Let the two double roots be a and b

=> x^4 - 16x^3 + 94x^2 + px + q = (x - a)^2 * (x - b)^2
=> x^4 - 16x^3 + 94x^2 + px + q = (x^2 - 2ax + a^2) * (x^2 - 2bx + b^2)
=> x^4 - 16x^3 + 94x^2 + px + q = x^4 - 2(a + b)x^3 + (a^2 + b^2 + 4ab)x^2 - 2ab(a + b)x + (ab)^2

Comparing coefficients on both sides,
8 = a + b
94 = a^2 + b^2 + 4ab
- 2ab(a + b) = p and
(ab)^2 = q

From the first two equations,
2ab
= (a^2 + b^2 + 4ab) - (a + b)^2
= 94 - 8^2
= 30

and from the next two equations,
p + q
= - 2ab(a + b) + (ab)^2
= - (30) * (8) + (15)^2
= - 240 + 225
= - 15.
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To find the values of p and q separately,
a + b = 8 and ab = 15
=> a - b = √[(a+b)^2 - 4ab] = 2
Solving, a = 5 and b = 3
 => p = - 2ab(a+b) = - 30*8 = - 240
and q = (ab)^2 = 15^2 = 225.
and the equal roots are 3 and 5.
(x - 5)^2 * (x - 3)^2
= (x^2 - 10x + 25) * (x^2 - 6x + 9)
= x^4 - 16x^3 + 94x^2 - 240x + 225.

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