Question

PLEASE HELP!! ASAP!!! 30 POINTS AND BRAINLIEST!!! PLEASE ANSWER!!

Answer 1

Part 1) x = 0 and x = two thirds

Part 2) x = −3; x = 0 is an extraneous solution

Part 3) (−∞, −3) ∪ (−3, ∞)

Answer 2

Answer:

Part 1) x = 0 and x = two thirds

Part 2) x = −3; x = 0 is an extraneous solution

Part 3) (−∞, −3) ∪ (−3, ∞)

Step-by-step explanation:

Part 1) we have

$\frac{x}{4}=\frac{x^{2}}{x+2}$

Solve for x

Multiply by 4(x+2) both sides to remove the fractions

$x(x+2)=4x^2$

$x^2+2x=4x^2\\3x^2-2x=0$

Solve the quadratic equation

Factor 3

$3(x^2-(2/3)x)=0$

Complete the square

$3(x^2-(2/3)x+4/36)=0+1/3$

Rewrite as perfect squares

$3(x-(1/3))^2=1/3$

$(x-(1/3))^2=1/9$

square root both sides

$(x-(1/3))=(+/-)1/3$

$x=(1/3)(+/-)1/3$

$x=(1/3)(+)1/3=2/3$

$x=(1/3)(-)1/3=0$

Part 2) we have

$\frac{3}{x}=\frac{4x+3}{x^{2}}$

Multiply by x^2 both sides to remove fractions

$3x=4x+3$

$x=-3$

Part 3) we have

$f(x)=\frac{x-1}{x+3}$

we know that

The denominator cannot be equal to zero

so

The value of x cannot be equal to -3

so

The domain of f(x) is all real numbers except the value of x=-3

therefore

(−∞, −3) ∪ (−3, ∞)