P= 2x^3-4x^2 +t^2* x -2

From Viete's rule we know that:

x1+x2+x3 = -b/a = 4/2 =2......(1)

x1x2+x1x3+x2x3= c/a = t^2/2........(2)

x1x2x3= -d/a= 1.......(3)

But we have x1+x2=1

==> substitute in (1):

1+ x3= 2

==> x3= 1

Now substitute x3=1 in (2)

x1x2+ x1 + x2= t^2/2

Now we know that x1+ x2 = 1

and x1x2x3= 1

but x3=1 ==> x1x2= 1

1+ 1= t^2/2

2= t^2/2

Multiply by 2:

4= t^2

==> t= +-2

P(x) = 2x^3-4x^2+t^2*x-2

x1+x2 =1.

Solution:

By the relation between the roots and cooefficients,

x1+x2+x3 =- (coeeficient of x^2/coefficient of x^3) = -(-4)/2 = 2.

x1x2+x2x3+x1x2 = x1(x2+x3)+x1x2= coefficient of x/coefficient of x^3 = t/2

x1x2x3 = -constant term / coefficient of x^3 = -(-2/2) =

Therefore x3 = 2- (x1+x2) =2-1 = 1, as given that x1+x2 =1

So the coefficient x/coefficient of x^3 = t/2 = x1(x2+x3)+x1x2 = 1*1 + (x1*x2*x3/x1)

= 1+ (x1x2x3)/(x1) = 1-{(-2)/2}/1 = 2

t/2 =2 .

t=4.

We notice that the polynomial has 3 roots, since it is of 3rd degree.

We'll use the first Viete's relation to calculate the 3rd root:

x1+x2+x3 = -(-4/2)

But, from enunciation, x1+x2 = 1.

1 + x3 = 2

We'll add -1 both sides:

x3 = 2-1

x3 = 1

Now, we'll use the second Viete's relation:

x1*x2+x1*x3+x2*x3 = t^2/2

We'll factorize and we'll get:

x3(x1+x2) + x1*x2 = t^2/2

But x1+x2 = 1 and x3 = 1

**x1*x2 + 1 = t^2/2 (1)**

We'll calculate x1*x2 from the 3rd Viete's relation:

x1*x2*x3 = -(-2/2)

x1*x2*1 = 1

x1*x2 = 1 (2)

We'll substitute (2) in (1) and we'll get:

1+1 = t^2/2

2 = t^2/2

We'll multiply by 2 both sides:

4 = t^2

**t1 = 2**

**t2 = -2**