What is the number of distinct possible rational roots of p(x)=2x2+7x+6?
I need help on how to solve this
3 answers
its meant to say p(x)=2x^2+7x+6
well a quadratic can have two roots but let's see
a = 2
b = 7
c = 6
x = [-b+/- sqrt(b^2-4ac)]/2a
if either is to be rational
sqrt(b^2-4ac) must be rational
sqrt (49 - 48) = sqrt 1 = 1
so ok, this will have 2 rational roots
they are
(-7 +1)/4 and(-7-1)/4
-3/2 and -2
===============
check
(x+ 3/2)(x+2) = x^2 + 7/2 x + 6/2 = 0
or 2 x^2 + 7 x + 6 = 0 ok
a = 2
b = 7
c = 6
x = [-b+/- sqrt(b^2-4ac)]/2a
if either is to be rational
sqrt(b^2-4ac) must be rational
sqrt (49 - 48) = sqrt 1 = 1
so ok, this will have 2 rational roots
they are
(-7 +1)/4 and(-7-1)/4
-3/2 and -2
===============
check
(x+ 3/2)(x+2) = x^2 + 7/2 x + 6/2 = 0
or 2 x^2 + 7 x + 6 = 0 ok
thank you
8 answers