Question

What are the approximate solutions of `5x^2 − 7x = 1` rounded to the nearest hundredth?

Answer 1

Subtracting `1` from both sides we get:

`5x^2-7x-1 = 0`

This is of the form `ax^2+bx+c = 0`, with `a = 5`, `b = -7` and `c = -1`.

The general formula for roots of such a quadratic gives us:

`x = (-b +- sqrt(b^2-4ac)) / (2a)`

`= (7+-sqrt((-7)^2-(4xx5xx-1)))/(2xx5)`

`= (7+-sqrt(69))/10`

`=0.7 +- sqrt(69)/10`

What is a good approximation for `sqrt(69)`?

We could punch it into a calculator, but let's do it by hand instead using Newton-Raphson:

`8^2 = 64`, so `8` seems like a good first approximation.

Then iterate using the formula:

`a_(n+1) = (a_n^2+69)/(2a_n)`

Let `a_0=8`

`a_1 = (64+69)/16 = 133/16 = 8.3125`

This is almost certainly good enough for the accuracy requested.

So `sqrt(69)/10 ~= 8.3 / 10 = 0.83`

`x ~= 0.7 +- 0.83`

That is `x ~= 1.53` or `x ~= -0.13`

Answer 2

Rewrite `5x^2-7x=1` in the standard form of `ax^2+bx+c = 0`
giving
`5x^2-7x-1 = 0`
then use the Quadratic Formula for roots:
`x = (-b+-sqrt(b^2-4ac))/(2a)`

In this case
`x = (7+-sqrt(49+20))/10`

Using a calculator:
`sqrt(69) = 8.306624` (approx.)

So
`x= 15.306624/10 = 1.53` (rounded to nearest hundredth)
or
`x = -1.306624/10 = -0.13` (rounded to the nearest hundredth)