Question

3.) \\(\sqrt{m - 1} + 5 = m - 2\\)

Explanation:

Step1: Isolate the square root

Subtract 5 from both sides of the equation $\sqrt{m - 1}+5 = m - 2$ to get $\sqrt{m - 1}=m - 2 - 5$, which simplifies to $\sqrt{m - 1}=m - 7$.

Step2: Square both sides

Square both sides of the equation $\sqrt{m - 1}=m - 7$ to eliminate the square root. We get $(\sqrt{m - 1})^2=(m - 7)^2$, which simplifies to $m - 1=m^2 - 14m + 49$.

Step3: Rearrange into quadratic equation

Rearrange the equation $m - 1=m^2 - 14m + 49$ to standard quadratic form $ax^2+bx + c = 0$. Subtract $m$ and add 1 to both sides: $0=m^2 - 14m - m+49 + 1$, so $m^2 - 15m + 50 = 0$.

Step4: Solve the quadratic equation

Factor the quadratic equation $m^2 - 15m + 50 = 0$. We need two numbers that multiply to 50 and add to - 15. The numbers are - 5 and - 10. So, $(m - 5)(m - 10)=0$. Setting each factor equal to zero gives $m - 5 = 0$ or $m - 10 = 0$, so $m = 5$ or $m = 10$.

Step5: Check for extraneous solutions

• For $m = 5$: Substitute into the original equation $\sqrt{5 - 1}+5=5 - 2$. $\sqrt{4}+5 = 3$, $2 + 5=7

eq3$. So, $m = 5$ is an extraneous solution.

• For $m = 10$: Substitute into the original equation $\sqrt{10 - 1}+5=10 - 2$. $\sqrt{9}+5 = 8$, $3 + 5 = 8$. So, $m = 10$ is a valid solution.

Answer:

$m = 10$