Quadratic & Zeros

Question

Find the zeros of a quadratic A and B, such that A+B=4 and *fractions*

(1/b) + (1/a) = 9

Answer 1

If A and B are zeroes of some quadratic, then (x - A)(x - B) = 0.

x² - Ax - Bx + AB = 0

x² - x(A + B) + AB = 0

Since A+B= 4, the equation becomes:

x² - 4x + AB = 0

Since 1/A + 1/B = 9, multiplying every term by AB gives:

B + A = 9AB

4 = 9AB (since A + B = 4)

And AB = 4/9.

Finally we have:

x² - 4x + 4/9 = 0

x² - 4x = -4/9

x² - 4x + 4 = -4/9 + 4 (completing the square)

(x - 2)² = 32/9

x - 2 = ±4√(2)/3

x = 2 ± 4√(2)/3

x = [6 ± 4√2]/3

A = [6 + 4√2]/3

B = [6 - 4√2]/3

A + B = 12/3 = 4

1/A = 1/[6+4√2]/3 = 3 / [6+4√2] = (9 + 6√2)/2

1/B = (9 - 6√2)/2

1/A + 1/B = 18/2 = 9

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