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

May 25, 2015

Subtracting $1$ from both sides we get:

$5 {x}^{2} - 7 x - 1 = 0$

This is of the form $a {x}^{2} + b x + c = 0$, with $a = 5$, $b = - 7$ and $c = - 1$.

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{7 \pm \sqrt{{\left(- 7\right)}^{2} - \left(4 \times 5 \times - 1\right)}}{2 \times 5}$

$= \frac{7 \pm \sqrt{69}}{10}$

$= 0.7 \pm \frac{\sqrt{69}}{10}$

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:

${a}_{n + 1} = \frac{{a}_{n}^{2} + 69}{2 {a}_{n}}$

Let ${a}_{0} = 8$

${a}_{1} = \frac{64 + 69}{16} = \frac{133}{16} = 8.3125$

This is almost certainly good enough for the accuracy requested.

So $\frac{\sqrt{69}}{10} \cong \frac{8.3}{10} = 0.83$

$x \cong 0.7 \pm 0.83$

That is $x \cong 1.53$ or $x \cong - 0.13$

May 25, 2015

Rewrite $5 {x}^{2} - 7 x = 1$ in the standard form of $a {x}^{2} + b x + c = 0$
giving
$5 {x}^{2} - 7 x - 1 = 0$
then use the Quadratic Formula for roots:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case
$x = \frac{7 \pm \sqrt{49 + 20}}{10}$

Using a calculator:
$\sqrt{69} = 8.306624$ (approx.)

So
$x = \frac{15.306624}{10} = 1.53$ (rounded to nearest hundredth)
or
$x = - \frac{1.306624}{10} = - 0.13$ (rounded to the nearest hundredth)