How do you find the zeros, real and imaginary, of y=13x^2-8x-9 using the quadratic formula?

Dec 7, 2015

$x \approx 0 \cdot 706 \text{ or } 0 \cdot 090$ to 3 decimal places

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Exact values:

$x = \frac{8 + \sqrt{107}}{26} \text{ or } \frac{8 - \sqrt{107}}{26}$

Thus: $x \in \mathbb{R}$

Explanation:

"What are the values of x if y=0. Find both real and any complex number solutions."

Using standard form: $a {x}^{2} + b x + c = 0$

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

$a = 13$
$b = - 8$
$c = - 9$

Giving: $\textcolor{w h i t e}{\ldots} x = \frac{8 \pm \sqrt{{8}^{2} - 4 \left(13\right) \left(- 9\right)}}{2 \left(13\right)}$

$\textcolor{w h i t e}{\ldots} x = \frac{8 \pm \sqrt{64 + 43}}{26}$

$\textcolor{w h i t e}{\ldots} x = \frac{8 \pm \sqrt{107}}{26}$

107 is a prime number so it may not be split into factors

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If you want approximate solutions we have:

$x \approx 0 \cdot 706 \text{ or } 0 \cdot 090$ to 3 decimal places

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Exact values:

$x = \frac{8 + \sqrt{107}}{26} \text{ or } \frac{8 - \sqrt{107}}{26}$

Thus: $x \in \mathbb{R}$