Question

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

Answer 1

`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`

Explanation:

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`