What are the approximate solutions of $5 x^{2} - 7 x = 1$ $5x^2 − 7x = 1$ rounded to the nearest hundredth?

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What are the approximate solutions of $5 x^{2} - 7 x = 1$ $5x^2 − 7x = 1$ rounded to the nearest hundredth?

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Answer 1 · 2015-05-25T12:52:10

Subtracting $1$ $1$ from both sides we get:

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

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

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2-4ac)) / (2a)$

$= \frac{7 \pm \sqrt{{(- 7)}^{2} - (4 \times 5 \times - 1)}}{2 \times 5}$ $= (7+-sqrt((-7)^2-(4xx5xx-1)))/(2xx5)$

$= \frac{7 \pm \sqrt{69}}{10}$ $= (7+-sqrt(69))/10$

$= 0.7 \pm \frac{\sqrt{69}}{10}$ $=0.7 +- sqrt(69)/10$

What is a good approximation for $\sqrt{69}$ $sqrt(69)$ ?

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

$8^{2} = 64$ $8^2 = 64$ , so $8$ $8$ seems like a good first approximation.

Then iterate using the formula:

$a_{n + 1} = \frac{a_{n}^{2} + 69}{2 a_{n}}$ $a_(n+1) = (a_n^2+69)/(2a_n)$

Let $a_{0} = 8$ $a_0=8$

$a_{1} = \frac{64 + 69}{16} = \frac{133}{16} = 8.3125$ $a_1 = (64+69)/16 = 133/16 = 8.3125$

This is almost certainly good enough for the accuracy requested.

So $\frac{\sqrt{69}}{10} ≅ \frac{8.3}{10} = 0.83$ $sqrt(69)/10 ~= 8.3 / 10 = 0.83$

$x ≅ 0.7 \pm 0.83$ $x ~= 0.7 +- 0.83$

That is $x ≅ 1.53$ $x ~= 1.53$ or $x ≅ - 0.13$ $x ~= -0.13$

Answer 2 · 2015-05-25T12:55:40

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

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

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

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

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What are the approximate solutions of $5 x^{2} - 7 x = 1$ $5x^2 − 7x = 1$ rounded to the nearest hundredth?

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What are the approximate solutions of 5x^2 − 7x = 15x2−7x=15x^2 − 7x = 1 rounded to the nearest hundredth?

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What are the approximate solutions of $5 x^{2} - 7 x = 1$ $5x^2 − 7x = 1$ rounded to the nearest hundredth?