# Understanding algebraic method of successive approximations to solve quadratic equations

I have straight away copied and pasted a worked example from john birds higher engineering mathematics {page 80 problem 4} which reads as follows.

Use an algebraic method of successive approximations to determine the value of the negative root of the quadratic equation: $4x^2 −6x −7=0$ correct to 3 significant figures. Check the value of the root by using the quadratic formula.

The solution goes like this.

A first estimate of the values of the roots is made by using the functional notation method $f(x) = 4x2 − 6x − 7$,

$f(0) = 4(0)2 − 6(0) − 7 = −7$

$f(−1) = 4(−1)2 − 6(−1) − 7 = 3$

These results showthat the negative root lies between 0 and −1, since the value of $f(x)$ changes sign between $f(0)$ and $f(−1)$ (see Section 9.1). The procedure given above for the root lying between 0 and −1 is followed.

What i'am not able to understand is this.

**First approximation

(a) Let a first approximation be such that it divides the interval 0 to −1 in the ratio of −7 to 3, i.e. let $x_{1}=−0.7$.**

What do they mean when they say "Let the first approx be such that it divides the interval 0 to -1 in the ratio of -7 to 3"?

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Answer in the initial spirit (supposing proportionality) : $$\begin{array} {c|ccccc} x&-1&&x_0&&0\\ \hline\\ f(x)&3&&0&&-7\\ \end{array}$$

$x_0$ must be at a place in $(-1,0)$ such that the ratio $−7$ to $3$ (on both sides of $f(x)=0$) is respected. Since the increase in $x$ is $1$ while the increase in $f(x)$ is $-10$ this means that $3$ and $-7$ must be replaced by an offset around $x_0$ of $-0.3$ and $0.7$ respectively getting $x_0-0.3=-1$.

(I started from $x=-1$ while they started from $0$ but the principle is the same!)

Visual answer (we want $x_0$ starting from $-1$) : $$\begin{array} {c|lccccc} &|\rightarrow&&+1&&\rightarrow|\\ &&\Delta_x=0.3&&&\\ x&-1&&x_0&&0\\ \hline\\ f(x)&3&&0&&-7\\ &&\Delta_l=-3&&\Delta_r=-7&\\ &|\rightarrow&&-10&&\rightarrow|\\ \end{array}$$ Observe that $\ \frac {\Delta_l}{\Delta_l+\Delta_r}=\frac {-3}{-10}=0.3\$ so that $\ \frac {\Delta_x}{1}=0.3\$ and $\ x_0=-1+\Delta_x=-0.7$

Text : Suppose that we move in a straight line from $f(-1)=3$ to $f(0)=-7$ (this is an interpolation!) then $x$ will progress of $\ 1=0-(-1)\$ while $f(x)$ progresses of $\ -10=-7-(3)$. If you want $f(x_0)=0$ then $f(x)$ must progress of $-3=0-3$ while $x$ will progress of $1\cdot \frac {-3}{-10}=0.3$ to get to $-1+0.3=-0.7$.

Algebraic : Suppose that $f$ is linear : $f(x)=ax+b$ with $f(-1)=3$ and $f(0)=-7$ then :

$-a+b=3$ and $b=-7$ so that (getting $a$ and $b$) we obtain $f(x)=-10x-7$.
So that $f(x_0)=0$ for $x_0=?$

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Continue this interpolation process, straddling the root all the time, and you have a very effective root finder ;-) – vonbrand Feb 1 '13 at 23:53

The short answer: pretend $f(x)$ is a linear function. Draw the line between $(-1, 3)$ and $(0,7)$. Take the place where it crosses the $x$-axis as a "guess" for the root of $f(x)$.

You could just take the average of $0$ and $-1$ as your next guess, but this method will often converge faster.

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You found that $f(-1)=3$ and $f(0)=-7$. So there is a root $r$ between $x=-1$ and $x=0$.

But we can do a little more. Note that $f(-1)$ is closer to $0$ than $f(0)$ is. So $r$ should be closer to $-1$ than it is to $0$.

How much closer? There is a gap of $10$ between $3$ and $-7$. So one would guess that the location of $r$ is roughly as follows. Divide the interval $[-1,0]$ into $10$ equal parts. Then $r$ should be about $3$ parts away from $-1$, and about $7$ parts away from $0$. Each part is of length $0.1$. So $r$ should be at about $-1+3(0.1)=-0.7$.

Remark: The justification depends on the fact that under magnification, a small section of a quadratic curve looks pretty much like a straight line. If we pretend we are on that straight line, the crossing point would be exactly at $-0.7$. Of course the quadratic between $-1$ and $0$ has some curviness, so $-0.7$ is only an approximation, but it is a good approximation.

What we are doing is called linear interpolation.

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