1. If the equation formed by decreasing each root of \( a x^{2}+b x+c=0 \) by ' 1 ' is \( 2 x^{2}+8 x+2=0 \), the values of \( \frac{a}{b} \) and \( \frac{b}{c} \).

To solve this problem, let's denote the roots of the original quadratic equation \( ax^2 + bx + c = 0 \) as \( r_1 \) and \( r_2 \). According to the problem, the new roots after decreasing each root by 1 are \( r_1 - 1 \) and \( r_2 - 1 \). Given that these new roots correspond to the quadratic equation \( 2x^2 + 8x + 2 = 0 \), we can find the sum and product of the roots from this equation. From Vieta's formulas, we know: - The sum of the new roots \( (r_1 - 1) + (r_2 - 1) = -\frac{b}{a} - 2 \) - The product of the new roots \( (r_1 - 1)(r_2 - 1) = \frac{c}{a} - \frac{b}{a} + 1 \) For the equation \( 2x^2 + 8x + 2 = 0 \): - The sum of the roots \( -(8/2) = -4 \) - The product of the roots \( 2/2 = 1 \) From the sum of the new roots, we set up the equation: \[ r_1 + r_2 - 2 = -4 \implies r_1 + r_2 = -2 \] From the product of the new roots, we have: \[ (r_1 - 1)(r_2 - 1) = r_1r_2 - (r_1 + r_2) + 1 = 1 \] Substituting \( r_1 + r_2 = -2 \) into the product equation gives: \[ r_1r_2 + 2 + 1 = 1 \implies r_1r_2 = -2 \] Now, we apply Vieta's relationships back: For the original equation \( ax^2 + bx + c = 0 \): 1. \( r_1 + r_2 = -\frac{b}{a} \) 2. \( r_1r_2 = \frac{c}{a} \) We already have \( r_1 + r_2 = -2 \) and \( r_1r_2 = -2 \): Thus, \(-\frac{b}{a} = -2 \implies \frac{b}{a} = 2\) which gives us \(\frac{a}{b} = \frac{1}{2}\). Additionally, we have: \(\frac{c}{a} = -2 \implies \frac{b}{c} = \frac{b/a}{c/a} = \frac{2}{-2} = -1\). In conclusion: \[ \frac{a}{b} = \frac{1}{2}, \quad \frac{b}{c} = -1. \]