Question

given that the zeros of the cubic polynomial x3-6x2+3x+10 are of the form a a+b a+2b for some positive real numbers, then value of a and b is​

Answer 1

Answer:

Step-by-step explanation:

Given:

${x}^{3} - 6 {x}^{2} + 3x +10$

Zeroes of cubic polynomial is in the form a,a+b and a+2b .

To find: Find the value of a and b.

Solution:We know that zeroes of cubic polynomial and coefficients of cubic polynomial have some relation.

If $\alpha ,\beta ,\gamma$

are the zeroes of cubic polynomial

$a{x}^{3} + b {x}^{2} + cx + d$

then

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} \\ \\ \alpha \beta \gamma = \frac{ - d}{a}$

So, here zeros of polynomial are a,a+b and a+2bapply the relation

$a + a + b + a + 2b = 6 \\ \\ or \\ \\ 3a + 3b = 6 \\ \\ or \\ \\\bold{ a + b = 2}...eq1$

From last relation

$a(a + b)(a + 2b) = -10 \\ \\ a(2)(a+b+b ) = -10 \\ \\2a(2+b)=-10\\\\\text{put value from eq1}\\\\ 2a (2+2 - a) = -10 \\ \\ 2a (4 - a) = -10 \\ \\or\\\\a(4-a)=-5\\ \\ 4a - {a}^{2} +5 = 0 \\ \\ or \\ \\ {a}^{2} -4a -5 = 0 \\ \\ {a}^{2} - 5a + a -5 = 0 \\ \\ a(a - 5) +1(a - 5) = 0 \\ \\ (a - 5)(a +1) = 0 \\ \\ either \\ \\ a = 5\\ \\ or \\ \\ a = -1$

ATQ values of a and b are some positive real values.

So discard negative value of a.Put a=5 in eq1

$5 + b = 2 \\ \\ b = - 3$

taking a positive will leads to negative value of b and vice versa.

If a = 5,b = -3If a = -1,b = 3

Final answer:

No such real positive values of a and b are exists.

Hope it helps you.

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