Question

Solve the given equations.

9a. (square root of 5y+1)+4=0

9b. 1+(square root of x+1)=(square root of 2x+3)

Answer 1

Question 22488

9a. $\sqrt{5y + 1} + 4 = 0$ , $\sqrt{5y + 1} = -4$ . Range of square root on the left: $y \geq -\frac{1}{5}$ .

Exponentiating both sides of equation, obtain $5y + 1 = 16, y = \frac{15}{5} = 3$ . This solution satisfies the domain, so this is the solution of equation.

9b. $1 + \sqrt{x + 1} = \sqrt{2x + 3}$ . First, find domains of squares: $x \geq -1, x \geq \frac{-3}{2} \Rightarrow x \geq -1$ .

Exponentiating both sides of equation, obtain $1 + x + 1 + 2\sqrt{x + 1} = 2x + 3, 2\sqrt{x + 1} = x + 1, 4(x + 1) = x^2 + 2x + 1 \Rightarrow x_{1,2} = -1; 3$ . Both solutions satisfy domains of squares, hence the solution of equation is $x = -1; x = 3$ .