Question

question
solve for all possible values of x.
$sqrt{29 - 5x} = x - 1$

Explanation:

Step1: Square both sides to eliminate root

$$(\sqrt{29-5x})^2=(x-1)^2$$
$$29-5x=x^2-2x+1$$

Step2: Rearrange to quadratic form

$$x^2-2x+1+5x-29=0$$
$$x^2+3x-28=0$$

Step3: Factor the quadratic equation

$$(x+7)(x-4)=0$$

Step4: Find potential solutions

$x+7=0 \implies x=-7$; $x-4=0 \implies x=4$

Step5: Verify solutions in original equation

For $x=-7$: $\sqrt{29-5(-7)}=\sqrt{29+35}=\sqrt{64}=8$, and $-7-1=-8$. $8
eq-8$, so $x=-7$ is extraneous.
For $x=4$: $\sqrt{29-5(4)}=\sqrt{29-20}=\sqrt{9}=3$, and $4-1=3$. $3=3$, so $x=4$ is valid.

Answer:

$x=4$