How do you solve #2x^2-3x-9=0#?

1 Answer
George C.
Jul 13, 2017

#x=-3/2" "# or #" "x=3#

Explanation:

Given:

#2x^2-3x-9=0#

Note that this quadratic is in the standard form:

#ax^2+bx+c=0#

with #a=2#, #b=-3# and #c=-9#.

Its discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = (color(blue)(-3))^2-4(color(blue)(2))(color(blue)(-9)) = 9+72 = 81 = 9^2#

Since this is positive and a perfect square, we can deduce that this quadratic has rational zeros that we can find by factoring with integer coefficients...

Use an AC method:

Look for a pair of factors of #AC=2*9=18# which differ by #B=3#.

The pair #6, 3# works in that #6*3=18# and #6-3=3#.

Use this pair to split the middle term and factor by grouping:

#0 = 2x^2-3x-9#

#color(white)(0) = (2x^2-6x)+(3x-9)#

#color(white)(0) = 2x(x-3)+3(x-3)#

#color(white)(0) = (2x+3)(x-3)#

Hence zeros:

#x=-3/2" "# or #" "x=3#

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