Question

Find the value of P such that the difference of the roots of the equation x^2 – Px + 8 = 0 is 2.

Answer 1

Answer:

Step-by-step explanation:

Let's take roots of this equation as a and b.

Then roots addition = a+b=-p/1=-p

roots multiplication =ab=8

But here difference of the roots=a-b=2

Then 8/b-b=2

8-b^2=2b

b^2+2b-8=0

(b+4)(b-2)=0

b=-4or b=2

So; roots addition -roots difference

a+b-a+b= -p-2

2b=(-p-2)

p=-2b-2

If b=-4 then p=-2*(-4)-2

=8-2

=6

If b=2 then p=-2*2-2

=-4-2

=(-6)

Answer 2

Step-by-step explanation:

Given equation :

x² + px + 8 = 0 .....(i)

Let, α and β are roots of (i).

Then,

α + β = - p .....(ii)

and

αβ = 8 .....(iii)

Given :

α - β = 2

=> (α - β)² = 2²

=> (α + β)² - 4αβ = 4

=> p² - (4 × 8) = 4

=> p² - 32 = 4

=> p² = 32 + 4 = 36 = 6²

=> p = ± 6

So, the required values of p are

p = ± 6.