Question

1. \\(sqrt{41 - 8a} + 7 = a\\)

Explanation:

Step1: Isolate the radical term

Subtract 7 from both sides to get the square root alone.
$\sqrt{41-8a} = a - 7$

Step2: Square both sides

Eliminate the square root by squaring each side.
$(\sqrt{41-8a})^2 = (a - 7)^2$
$41 - 8a = a^2 - 14a + 49$

Step3: Rearrange to quadratic form

Move all terms to one side to set the equation to 0.
$a^2 - 14a + 49 - 41 + 8a = 0$
$a^2 - 6a + 8 = 0$

Step4: Factor the quadratic

Break the quadratic into binomial factors.
$(a - 2)(a - 4) = 0$

Step5: Solve for a

Set each factor equal to 0 and solve.
$a - 2 = 0 \implies a=2$; $a - 4 = 0 \implies a=4$

Step6: Check for extraneous solutions

Substitute solutions back into the original equation.
For $a=2$: $\sqrt{41-8(2)} +7 = \sqrt{25}+7=5+7=12
eq 2$ (invalid)
For $a=4$: $\sqrt{41-8(4)} +7 = \sqrt{9}+7=3+7=10
eq 4$ (invalid)
Wait, recheck squaring step:
Wait, original after step1: $\sqrt{41-8a}=a-7$. The right-hand side must be non-negative, so $a-7\geq0 \implies a\geq7$.
Our solutions are 2 and 4, both less than 7, so they are extraneous. Wait, recheck quadratic expansion:
$(a-7)^2=a^2-14a+49$, correct. $41-8a=a^2-14a+49$ → $a^2-6a+8=0$, correct. Roots 2 and 4. But since $\sqrt{41-8a}\geq0$, $a-7\geq0 \implies a\geq7$. Substitute $a=7$: $\sqrt{41-56}+7=\sqrt{-15}+7$, not real. $a=8$: $\sqrt{41-64}+7=\sqrt{-23}+7$, not real. $a=1$: $\sqrt{41-8}+7=\sqrt{33}+7\approx5.744+7=12.744
eq1$. Wait, actually, there is no real solution? Wait no, wait did I rearrange wrong?
Original equation: $\sqrt{41-8a}+7=a$ → $\sqrt{41-8a}=a-7$. For the square root to be real, $41-8a\geq0 \implies a\leq\frac{41}{8}=5.125$. But $a-7\geq0 \implies a\geq7$. There is no overlap between $a\leq5.125$ and $a\geq7$, so no real solutions.

Answer:

There is no real solution for $a$.