Given that 1/x1+1/x2=2, what is t such as x^2-(t-2)x+t+1=0?

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Since x1 and x2 are the roots of the equation, we'll apply Viete's relations.

x1 + x2 = t - 2

x1*x2 = t + 1

But 1/x1 + 1/x2 = 2 <=> (x1+x2)/x1*x2 = 2 => x1 + x2 = 2x1*x2

t - 2 = 2t + 2

We'll isolate t to the left side:

t - 2t = 2 + 2

-t = 4

t = -4

**The value of t, such as 1/x1 + 1/x2 = 2, is: t = -4.**

It is given that 1/x1 + 1/x2 = 2.

t has to be determined such that x^2 - (t-2)x + t +1 = 0

1/x1 + 1/x2 = 2

=> (x1 + x2)/x1*x2 = 2

For a quadratic equation ax^2 + bx + c = 0 with roots x1 and x2,

c/a = x1*x2 and -b/a = x1 + x2

Here the equation is x^2 - (t-2)x + t +1 = 0

=> (t - 2)/(t + 1) = 2

=> t - 2 = 2t + 2

=> t = -4

**The required value of t = -4**

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