How do you solve the quadratic #2/(x+5)-x/(x-5)=1# using any method?

1 Answer
Douglas K.
Jan 5, 2017

Stipulate that #x !=+-5#. Multiply both sides by #(x - 5)(x + 5)#. Collect all the terms on one side. Factor or use the quadratic formula. Discard any restricted roots.

Explanation:

Stipulate that #x !=+-5#:

#2/(x + 5) - x/(x - 5) = 1; x !=+-5#

Multiply both sides by #(x - 5)(x + 5)#:

#2(x - 5) - x(x + 5) = x^2 - 25; x !=+-5#

#2x - 10 - x^2 - 5x = x^2 - 25; x !=+-5#

Collect all of the terms to one side:

#2x^2 + 3x - 15 = 0;x !=+-5#

Check the discriminant:

#d = b^2 -4(a)(c) = 3^2 - 4(2)(-15) = 129#

Drop the restriction, because, with 129 under the radical, there is no way to obtain a root equal to #+-5#.

Use the quadratic formula:

#x = (-b +-sqrt(d))/(2a)#

#x = (-3 +-sqrt(129))/(2(2))#

#x = {((-3 +sqrt(129))/(4)),((-3 -sqrt(129))/(4)):}#

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