Question

(03.05 MC) Solve the rational equation x divided by 2 equals x squared divided by quantity x minus 2 end quantity, and check for extraneous solutions. No solution x = 0 and x = −2 x = −2; x = 0 is an extraneous solution x = 0; x = −2 is an extraneous solution

Answer 1

Answer:

x = 0 and x = -2 are solutions of the given rational equation.

Step-by-step explanation:

We must solve the following rational equation:

$\frac{x}{2} = \frac{x^{2}}{x-2}$

Now we present the procedure:

1) $\frac{x}{2} = \frac{x^{2}}{x-2}$ Given

2) $x\cdot (x-2) = 2\cdot x^{2}$ Compatibility with multiplication/Existence of the multiplicative inverse/Definition of division/Modulative property.

3) $x^{2}-2\cdot x = 2\cdot x^{2}$ Distributive property/$a^{b}\cdot a^{c} = a^{b+c}$

4) $x^{2} + 2\cdot x = 0$ Compatibility with addition/Existence of the additive inverse/Modulative property/Reflexive property

5) $x \cdot (x+2) = 0$ Distributive property/$a^{b}\cdot a^{c} = a^{b+c}$

6) $x = 0\, \lor\, x = -2$ Result

Now we check the rational equation with each root:

x = 0

$\frac{x}{2} = \frac{x^{2}}{x-2}$

$\frac{0}{2} = \frac{0^{2}}{0-2}$

$0 = \frac{0}{-2}$

$0 = 0$

x = 0 is a solution of the rational equation.

x = -2

$\frac{x}{2} = \frac{x^{2}}{x-2}$

$\frac{-2}{2} = \frac{(-2)^{2}}{-2-2}$

$-1 = -1$

x = -2 is a solution of the rational equation.