Question

learning goal from 11.3: i can solve radical equations in one variable and determine extraneous solutions. i can explain how extraneous solutions may arise from radical equations. identify the solutions and any extraneous solutions of the following equations. (1 point each) 15. $sqrt{2x + 5}=sqrt{3x - 1}$ 16. $sqrt{5x - 4}-6 = 0$ 17. $sqrt{-4x + 12}=x$

Explanation:

Step1: Square both sides of the equation

For $\sqrt{2x + 5}=\sqrt{3x - 1}$, squaring both sides gives $2x + 5=3x - 1$.

Step2: Solve for $x$

Subtract $2x$ from both sides: $5=x - 1$. Then add 1 to both sides, so $x = 6$.

Step3: Check for extraneous solutions

For the left - hand side, when $x = 6$, $\sqrt{2\times6+5}=\sqrt{12 + 5}=\sqrt{17}$. For the right - hand side, $\sqrt{3\times6-1}=\sqrt{18 - 1}=\sqrt{17}$. So $x = 6$ is a valid solution.

For the equation $\sqrt{5x - 4}-6 = 0$:

Step1: Isolate the radical

Add 6 to both sides of the equation $\sqrt{5x - 4}-6 = 0$, we get $\sqrt{5x - 4}=6$.

Step2: Square both sides

Squaring both sides gives $5x-4 = 36$.

Step3: Solve for $x$

Add 4 to both sides: $5x=40$. Then divide both sides by 5, so $x = 8$.

Step4: Check for extraneous solutions

When $x = 8$, $\sqrt{5\times8-4}=\sqrt{40 - 4}=\sqrt{36}=6$. So $x = 8$ is a valid solution.

For the equation $\sqrt{-4x + 12}=x$:

Step1: Square both sides

Squaring both sides gives $-4x + 12=x^{2}$.

Step2: Rearrange to quadratic form

Rearrange to $x^{2}+4x - 12 = 0$.

Step3: Factor the quadratic

Factor as $(x + 6)(x - 2)=0$. So $x=-6$ or $x = 2$.

Step4: Check for extraneous solutions

When $x=-6$, the left - hand side is $\sqrt{-4\times(-6)+12}=\sqrt{24 + 12}=\sqrt{36}=6$, and the right - hand side is $-6$. So $x=-6$ is an extraneous solution.
When $x = 2$, the left - hand side is $\sqrt{-4\times2+12}=\sqrt{-8 + 12}=\sqrt{4}=2$, and the right - hand side is 2. So $x = 2$ is a valid solution.

Answer:

1. Solution: $x = 6$, No extraneous solutions.
2. Solution: $x = 8$, No extraneous solutions.
3. Solution: $x = 2$, Extraneous solution: $x=-6$