a) Step 1: Identify the sum and product of the roots of the given quadratic equation. The given quadratic equation is $2x^2 - 5x + k = 0$. The roots are $x_1 = 1+\alpha$ and $x_2 = 1+\beta$. For a quadratic equation $ax^2+bx+c=0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$. Sum of roots: $$(1+\alpha) + (1+\beta) = -\frac{-5}{2}$$ $$2 + \alpha + \beta = \frac{5}{2}$$ $$\alpha + \beta = \frac{5}{2} - 2$$ $$\alpha + \beta = \frac{1}{2}$$ Product of roots: $$(1+\alpha)(1+\beta) = \frac{k}{2}$$ $$1 + \alpha + \beta + \alpha\beta = \frac{k}{2}$$ Substitute $\alpha + \beta = \frac{1}{2}$: $$1 + \frac{1}{2} + \alpha\beta = \frac{k}{2}$$ $$\frac{3}{2} + \alpha\beta = \frac{k}{2}$$ Multiply by 2: $$3 + 2\alpha\beta = k$$ $$2\alpha\beta = k-3$$ $$\alpha\beta = \frac{k-3}{2}$$ Step 2: Find the sum and product of the roots of the new quadratic equation. The roots of the new quadratic equation are $y_1 = 1+\alpha^2$ and $y_2 = 1+\beta^2$. Let $S'$ be the sum of the new roots and $P'$ be the product of the new roots. Sum of new roots $S'$: $$S' = (1+\alpha^2) + (1+\beta^2)$$ $$S' = 2 + \alpha^2 + \beta^2$$ We know that $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$. Substitute the values of $\alpha+\beta$ and $\alpha\beta$: $$\alpha^2 + \beta^2 = \left(\frac{1}{2}\right)^2 - 2\left(\frac{k-3}{2}\right)$$ $$\alpha^2 + \beta^2 = \frac{1}{4} - (k-3)$$ $$\alpha^2 + \beta^2 = \frac{1}{4} - k + 3$$ $$\alpha^2 + \beta^2 = \frac{1+12}{4} - k$$ $$\alpha^2 + \beta^2 = \frac{13}{4} - k$$ Now substitute this back into the expression for $S'$: $$S' = 2 + \left