determine the nature of the roots of 2x squared - squareroot of 3x+8=0?

Question

The sum of the squares of three consecutive odd integers is 371. Find the integers. Find k so that (3k+1)x squared-(2k-1)x+5=0 has one root equal to 1.

Answer 1

Answer to the 3rd of your 3 questions:

„Find k so that (3k+1)x squared-(2k-1)x+5=0 has one root equal to 1.“

(3*k+1)*(x^2) - (2*k-1)*x + 5 = 0

One root have to be equal to 1, therefore the next equation is valid:

(3*k+1) - (2*k-1) + 5 = 0

From the above equation we may find the value of k:
3*k + 1 - 2*k + 1 + 5 = 0
k = -7

Is it clear?

(BTW the second root is equal to -0.25.)
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