Question

if alpha beta Gama are zeros of x^3+px^2+qx+2 such that alpha multiple by beta +1=0 find the value of 2p+q+5

Answer 1

X^3+px^2+qx+2=0
A+B+1=0
A+B=-1
A+B+Y=-2
-1×A=-2
A=2
A+B+Y=-p
A+B=-p-2----------(1)
AB+BY+YA=c/a
AB+2B+2A=c/a
-1+2 (A+B)=q
A+B=q+1/2---------(2)
Using (1)&(2)
-p-2=q+1/2
-2p-4=q+1
-2p-q-5=0
Hence,2p+q+5=0

Answer 2

Answer:

Value of 2p + q + 5 = 0

Step-by-step explanation:

If α, β and γ are three zeros of x³ + px² + qx + 2 then

(x - α)(x - β)(x - γ) = 0

(x² - αx - βx + αβ)(x - γ) = 0

x³ - αx² - βx² + αβx - γx² + αγx + βγx - αβγ = 0

x³ - x²(α + β + γ) + x(αβ + βγ + αγ) - αβγ = 0

Now we compare it with x³ + px² + qx + 2 = 0

Then  p = -(α + β + γ), q = (αβ + βγ + αγ) and αβγ = -2

Moreover this αβ + 1 = 0

Or αβ = -1

Since αβγ = -2 ⇒ (-1)γ = -2

⇒ γ = 2

Now p = -(α + β + 2),  q = (-1 + 2β + 2α)

Now we have to evaluate 2p + q + 5

-2(α + β + 2) + (-1 + 2β + 2α) + 5 = -2α - 2β - 4 - 1 + 2α + 2β + 5 = 0

Therefore, 2p + q + 5 = 0

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