The solution set of
[tex] \frac{x ^{2} }{x + 1} - 1 = \frac{1}{x + 1} [/tex]
is

A) {-1, 2}
B) {2}
C) {0, 1}
D) {1, -2}

[tex]\dfrac{x^2}{x+1}-1-\dfrac{1}{x+1}=0\\\\\dfrac{x^2-x-1-1}{x+1}=0\\\\\dfrac{(x-2)(x+1)}{x+1}=0\\\\x-2=0 \qquad\text{cancel like terms}\\\\x=2[/tex]

The quadratic has solutions x={-1, 2}, but x=-1 makes the original equation undefined. It is extraneous. The only valid solution is x = 2.

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Comment on this solution method

Often, subtracting one side of a rational equation so you get an equation of the form ( )/( ) = 0 will let you cancel common factors from numerator and denominator. This can help avoid any extraneous solutions. (You may still get an extraneous solution if a denominator factor appears at a higher power in the numerator. x³/x² = 0 reduces to x=0, but that is still extraneous.)

The solution set of [tex] \frac{x ^{2} }{x + 1} - 1 = \frac{1}{x + 1} [/tex]isA) {-1, 2}B) {2}C) {0, 1}D) {1, -2}​

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The solution set of
[tex] \frac{x ^{2} }{x + 1} - 1 = \frac{1}{x + 1} [/tex]
is

A) {-1, 2}
B) {2}
C) {0, 1}
D) {1, -2}