# Quadratic Formula expressed with Taylor's Theorem

I am having trouble solving the problem below. I think I understand the first part by just doing a taylor expansion of $f(\delta - a)$ where $a=0$ and the function equals $\sqrt{1-\delta}$. But I do not know how to do the second part of expressing the quadratic formula in terms of $\sqrt{1-\delta}$. Any help is appreciated.

Consider the quadratic equation $ax^{2} + bx +c = 0$, where $a=1, b=3, c= 10^{-11}$.

Show with Taylor's theorem that for any small real number $\delta$,

$\sqrt{1-\delta)} = 1 - \frac{1}{2}\delta + O(\delta^{2})$

And based on the formula above approximate $(-b \pm \sqrt{b^{2} - 4ac})/2a$ and estimate the two roots of the given quadratic equation analytically. You need to find proper values of $\delta$ based on $a, b, c$ so that your quadratic formula can be expressed in terms of $\sqrt{1-\delta}$.

We have $\sqrt{b^2-4ac}=b\sqrt{1-4ac/b^2}$. To approximate $\sqrt{1-4ac/b^2}$, you can use the Taylor approximation with $\delta=4ac/b^2$, which is a small number.
• How does that help me find the roots of the function $x^{2} + 3x + 10^{-11}$? – Vivek Feb 24 '15 at 4:24
• The quadratic formula gives you the two roots (as you wrote in your question), and by using this approximation with $a=1,b=3,c=10^{-11}$, you can approximate the roots. – TorsionSquid Feb 24 '15 at 4:26
• Yeah but I need to rewrite the quadratic formula in terms of $\sqrt{1-\delta}$ and then figure out the roots with the new formula, rather then directly plugging in $a, b, c$. How would I do that? – Vivek Feb 24 '15 at 4:29
• I explain in my answer how to write the square-root part of the quadratic formula to look like $\sqrt{1-\delta}$, so that the Taylor approximation can be used. – TorsionSquid Feb 24 '15 at 4:30