Question

lesson 11.3 checkpoint
once you have completed the above problems and checked your solutions, complete the lesson checkpoint below.
complete the lesson reflection above by circling your current understanding of the learning goal.
identify the solutions and any extraneous solutions of the following equations.

1. $sqrt{15x - 10}=sqrt{2x + 3}$
2. $sqrt{45 - 9x}-6 = 0$
3. $sqrt{x + 5}=x - 1$

Explanation:

Step1: Square both sides of equation 1

Square both sides of $\sqrt{15x - 10}=\sqrt{2x + 3}$ to get $15x-10 = 2x+3$.

Step2: Solve for $x$ in equation 1

Subtract $2x$ from both sides: $15x-2x-10=2x - 2x+3$, which simplifies to $13x-10 = 3$. Then add 10 to both sides: $13x-10 + 10=3 + 10$, so $13x=13$. Divide both sides by 13, we have $x = 1$.

Step3: Check for extraneous solutions in equation 1

Substitute $x = 1$ into the original equation. Left - hand side: $\sqrt{15\times1-10}=\sqrt{5}$, right - hand side: $\sqrt{2\times1 + 3}=\sqrt{5}$. So $x = 1$ is a valid solution.

Step4: Isolate the radical in equation 2

For $\sqrt{45-9x}-6 = 0$, add 6 to both sides to get $\sqrt{45-9x}=6$.

Step5: Square both sides of equation 2

Square both sides: $45-9x = 36$.

Step6: Solve for $x$ in equation 2

Subtract 45 from both sides: $-9x=36 - 45=-9$. Divide both sides by - 9, we get $x = 1$.

Step7: Check for extraneous solutions in equation 2

Substitute $x = 1$ into the original equation. $\sqrt{45-9\times1}-6=\sqrt{36}-6=6 - 6=0$. So $x = 1$ is a valid solution.

Step8: Square both sides of equation 3

Square both sides of $\sqrt{x + 5}=x - 1$ to get $x + 5=(x - 1)^2$. Expand the right - hand side: $x + 5=x^{2}-2x + 1$.

Step9: Rearrange to a quadratic equation in equation 3

Rearrange to $x^{2}-2x+1-x - 5=0$, which simplifies to $x^{2}-3x - 4=0$.

Step10: Factor the quadratic equation in equation 3

Factor $x^{2}-3x - 4$ as $(x - 4)(x+1)=0$.

Step11: Solve for $x$ in equation 3

Set each factor equal to zero: $x - 4=0$ gives $x = 4$, and $x+1=0$ gives $x=-1$.

Step12: Check for extraneous solutions in equation 3

For $x = 4$, left - hand side: $\sqrt{4 + 5}=3$, right - hand side: $4 - 1=3$. For $x=-1$, left - hand side: $\sqrt{-1 + 5}=2$, right - hand side: $-1-1=-2$. So $x=-1$ is an extraneous solution and $x = 4$ is a valid solution.

Answer:

1. Solution: $x = 1$, No extraneous solutions.
2. Solution: $x = 1$, No extraneous solutions.
3. Solution: $x = 4$, Extraneous solution: $x=-1$.