X^2-14x+ 58 = 0
A. (3+ 7i,3 - 7i)
B. (2+19i,2 - 19i)
c. (2+9i,2 - 9i)
D. (7+3i,7-3i)

9514 1404 393

 D.  7 +3i, 7 -3i

Step-by-step explanation:

When the leading coefficient of the quadratic is 1, as it is here, the sum of the roots is equal to the opposite of the coefficient of the x-term. So, we need the sum of roots to be -(-14) = 14.

In each case, the imaginary parts cancel when the roots are added. The real parts of the roots are the same, so the sum of roots is double the real part of one of the roots. In order for that to be 14, the real part of the root must be 7.

Only answer choice D matches this requirement.

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Algebraic solution

  x^2 -14x +58 = 0 . . . . given

  (x -7)^2 = -9 . . . . . . . subtract 9 and write as a square

  x = 7 ± 3i . . . . . . . . square root and add 7

Additional comment

When p and q are roots, (x-p) and (x-q) are factors. The expansion of the factored form is ...

  (x -p)(x -q) = x^2 -(p+q)x +pq = 0

That is, the constant term is the product of the roots, and the linear (x) term coefficient is the opposite of the sum of the roots. This can help you check quickly if an offered choice of roots is applicable to the given equation. This is also the basis of the thinking used to factor the equation.

how is that middle school

step-by-step explanation: