How do you solve #9x^2-5x=7#?

1 Answer
Aug 8, 2015

Answer:

Subtract #7# from both sides to get: #9x^2-5x-7 = 0#, then solve using the quadratic formula to get:

#x = (5+-sqrt(277))/18#

Explanation:

#9x^2-5x-7# is of the form #ax^2+bx+c#, with #a=9#, #b=-5# and #c=-7#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-5)^2-(4xx9xx-7) = 25+252#

#= 277#

Since #Delta > 0#, the quadratic equation has two distinct Real roots. Since #Delta# is not a perfect square (#277# is prime), those roots are irrational.

The roots are given by the quadratic formula:

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

#=(-b+-sqrt(Delta))/(2a)#

#=(5+-sqrt(277))/18#

Notice the discriminant #Delta# is the expression under the square root.

So if #Delta < 0# the square root is not Real and the quadratic has no Real roots - It has a conjugate pair of distinct complex roots.

If #Delta = 0# then there is one repeated Real root.

If #Delta > 0# (as in our example), there are two distinct Real roots. If in addition #Delta# is a perfect square, then those roots are rational.