Question

x^4+ax^3+bx^2-32x+16=0 has all roots positive find a and b?

Answer 1

you have to know the relationship between coefficient of n degree and roots of it .
Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n_2}x^{n-2}+.........+a_1x+a_0$
$\bold{\text{sum of roots}}=-\frac{a_{n-1}}{a_n}\\\\\bold{\text{sum product of roots , 2 at a times}}=\frac{a_{n-2}}{a_n}\\\\\bold{\text{sum product of roots ,3 at a times}}=-\frac{a_{n-3}}{a_n}\\\\so,\bold{\text{sum product of roots , n at a times }}=(-1)^n\frac{a_0}{a_n}$

Now, use it here,
Let α, β, γ, δ are the roots of x⁴ + ax³ + bx² -32x + 16 = 0
then, α + β + γ + δ = -a/1 = -a _____(1)
αβ + βγ + γδ + δα + δβ + γα = b/1 = b ______(2)
αβγ + βγδ + γδα + δαβ = -(-32)/1 = 32 ________(3)
αβγδ = 16 ________(4)

now, assume α = 2, β = 2 , γ = 2 and δ = 2
check it by putting these in equation (3) and (4)

Now, put the values of α, β, γ and δ in equations (1) and (2) and find the value of a and b .
α + β + γ + δ = -a
2 + 2 + 2 + 2 = -a ⇒ a = -8

αβ + βγ + γδ + γα + δα + δβ = b
2 × 2 + 2 × 2 + 2 × 2 + 2 × 2 + 2 × 2 + 2 × 2 = b
⇒4 + 4 + 4 + 4 + 4 + 4 = b
⇒ 24 = b

Hence, a = -8 and b = 24

Answer 2

Given : x⁴+ax³+bx²-32x+16=0  has all roots positive

To Find : a and b

Solution:

x⁴+ax³+bx²-32x+16=0

Let say  α₁ , α₂ ,  α₃  , α₄ are the roots

α₁α₂α₃α₄  = 16

α₁α₂α₃ +  α₁α₂α₄ +  α₁α₃α₄  +  α₂α₃α₄ = -(-32)/1  

=> 16/α₄ + 16/α₃  +16/α₂ + 16/α₁  = 32

=> 1/α₄ + 1/α₃  +1/α₂ + 1/α₁  =  2

=> AM of  1/α₄ , 1/α₃ , 1/α₂ ,  1/α₁   = 2/4 = 1/2

α₁α₂α₃α₄  = 16  

=> 1/ α₁α₂α₃α₄ = 1/16

=> GM of  1/α₄ , 1/α₃ , 1/α₂ ,  1/α₁  = 1/(16)^(1/4) =  1/2

AM = GM

Hence  1/α₄ = 1/α₃ = 1/α₂ =  1/α₁  

=>  1/α₄ + 1/α₃ + 1/α₂ +  1/α₁   = 2

=>  1/α₄ = 1/α₃ = 1/α₂ =  1/α₁   = 1/2

=> α₁ = α₂ = α₃  = α₄ = 2

Hence ( x - 2)⁴  = 0

=>  x⁴ - 8x³ + 24x² - 32x + 16  = 0

Hence a = - 8

  b = 24

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