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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