9514 1404 393
Answer:
t = ±√3
Explanation:
The first equation tells us ...
x^2 +tx +2 = (x -h)(x -k)
x^2 +tx +2 = x^2 -(h+k)x +hk
Comparing coefficients, we have the equations ...
t=-(h+k)
2 = hk
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The second equation tells us ...
2x^2 +mx +2m = 2(x -h/k)(x -k/h)
2x^2 +mx +2m = 2x^2 -2(h/k +k/h)x +2
Comparing coefficients, we have the equations ...
m = -2(h/k +k/h)
2m = 2
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Dividing the last equation by 2 gives ...
m = 1
Using that in the third equation, we have ...
1 = -2(h/k +k/h) = -2(h^2 +k^2)/hk
Using hk = 2, this gives us ...
h^2 +k^2 = -1
From the first equation, we know ...
t^2 = (-(h+k))^2 = h^2 +2hk +k^2 = (h^2 +k^2) +2(hk)
Substituting the values for these terms, we have ...
t^2 = -1 + 2(2) = 3
t = ±√3
Possible values of t are √3 and -√3.