Considering the quadratic equation:
$$ (x-1)(x-2) = x^2 - 3x + 2$$
which has exact roots, $x_1 = 1 \quad and \quad x_2 =2.$
Supposing that a numerical algorithm finds approximate roots equal to $ \hat x_1 =1.1 \quad and \quad \hat x_2=1.9,$
I am quite confused on how to proceed with this question.
Could anyone please show me or walk me through how to:
for each approximate root, find a new quadratic which is as close to the original quadratic, as possible, and for which the approximate root is the exact root. That is, for each approximate root, find a δ such that $ x^2 - 3x + 2 + δ$ has the approximate root as an exact root.
find the backward error and relative backward error for each approximate root
Let $p : \mathbb{R}^2 \rightarrow \mathbb{R}$ be given by $$p(x,\delta) = x^2 - 3x + 2 + \delta.$$ Then $p(1.1,\delta) = \delta - 0.09$ and $p(1.9,\delta)= \delta - 0.09$. Hence, the computed roots are the exact roots of the polynomial $q$ given by $$q(x) = p(x,0.09) = x^2 - 3x + 2.09.$$ The absolute value of the backward error for each root is $0.09$, the difference between the new constant term $2.09$ and the old value $2$. The absolute value of the relative backward error is just $0.09/2 = 0.045$.