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

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Answer 1 · 2015-12-07T16:45:09

$x~~ 0*706 " or " 0*090$ to 3 decimal places

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Exact values:

$x = (8+ sqrt(107))/26 " or " (8- sqrt(107))/26$

Thus: $x in RR$

To be mathematically correct your question should read as:

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

Using standard form: $ax^2+bx+c=0$

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

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

Giving: $color(white)(...)x=(8+-sqrt(8^2-4(13)(-9)))/(2(13))$

$color(white)(...)x=(8+-sqrt(64+43))/(26)$

$color(white)(...)x=(8+-sqrt(107))/(26)$

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

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If you want approximate solutions we have:

$x~~ 0*706 " or " 0*090$ to 3 decimal places

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

$x = (8+ sqrt(107))/26 " or " (8- sqrt(107))/26$

Thus: $x in RR$

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