# Math problem quadratic equation?

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1/x-1 + 1/x-2 + 1/x-3 = 0
asked Feb 4, 2013

## 1 Answer

+1 vote

1 / (x - 1) + 1 / (x - 2) + 1 / (x - 3) = 0

Rewrite the expression with common denominator.

[ (x - 2)(x - 3) + (x - 1)(x - 1) + (x - 1)(x - 2)] / (x - 1)(x - 2)(x - 3) = 0.

Multiply each side by (x - 1)(x - 2)(x - 3).

(x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2) = 0

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

x2 - 3x - 2x + 6 + x2 - 3x - 1x + 3 + x2 - 2x -1x + 2 = 0.

3x2 - 12x + 11 = 0.

Compare equation with standard from ax2 + bx + c = 0 and write the coefficients.

a = 3, b = -12, and c = 11.

Quadratic formula: x = [-b ± √(b2 - 4ac)] / 2a.

Substitute a = 3, b = -12, and c = 11 in the quadratic formula.

x = [-(-12) ± √((-12)2 - 4(3)(11))] / 2(3).

x = [12 ± √(144 - 132)] / 6.

x = [12 ± √(12)] / 6.

x = [12 ± √(4·3)] / 6

x = [12 ± 2√(3)] / 6

x = [12 + 2√(3)] / 6 or x  =[12 - 2√(3)] / 6

x = [12 + 2√(3)] / 6.

Take out common term 2.

x = 2 [6+√(3)] / 6.

x = [6+√3] / 3 = (6/3) + (√3/3) = 2 + 1/√3

Similarly x = [2 - 1/√3].

Therefore x = 2 + 1/√3 or 2 - 1/√3.

answered Feb 4, 2013