Question:

Step 1: Let the roots of $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$. Step 2: An equation whose roots are decreased by 1, i.e., $(\alpha-1)$ and $(\beta-1)$, can be obtained by replacing $x$ with $(x+1)$ in the original equation. Step 3: Substitute $x \to (x+1)$ into $ax^2 + bx + c = 0$: $$a(x+1)^2 + b(x+1) + c = 0$$ Step 4: Expand and group terms: $$a(x^2 + 2x + 1) + b(x + 1) + c = 0$$ $$ax^2 + 2ax + a + bx + b + c = 0$$ $$ax^2 + (2a + b)x + (a + b + c) = 0$$ Step 5: This transformed equation must be equivalent to $2x^2 + 8x + 2 = 0$. For two quadratic equations to have the same roots, their coefficients must be proportional. Thus, there exists a non-zero constant $k$ such that: $$a = 2k \quad (1)$$ $$2a + b = 8k \quad (2)$$ $$a + b + c = 2k \quad (3)$$ Step 6: Solve the system of equations for $a, b, c$ in terms of $k$. From (1), $a = 2k$. Substitute $a=2k$ into (2): $$2(2k) + b = 8k$$ $$4k + b = 8k$$ $$b = 4k$$ Substitute $a=2k$ and $b=4k$ into (3): $$2k + 4k + c = 2k$$ $$6k + c = 2k$$ $$c = -4k$$ Step 7: Determine the relationship between $a, b,$ and $c$. We have $a=2k$, $b=4k$, and $c=-4k$. Observe the relationship between $b$ and $c$: $$b = 4k$$ $$-c = -(-4k) = 4k$$ Therefore, $b = -c$.