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

1 Answer
Dec 7, 2015

Answer:

#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#