SOLUTION: 42.) Find b^2 � 4ac and the number of real solutions to each equation. 9m^2+16=24m

Question 164917This question is from textbook Elementary and Intermediate
: 42.) Find b^2 � 4ac and the number of real solutions to each equation.
9m^2+16=24m
This question is from textbook Elementary and Intermediate

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Find b^2 � 4ac and the number of real solutions to each equation.
9m^2+16=24m
.
First, convert your equation into the form of:
am^2 + bm + c = 0
.
Starting with:
9m^2+16=24m
subtracting 24m from both sides:
9m^2-24m+16=0
we "see" that:
a = 9
b = -24
c = 16
.
b^2 � 4ac
is called the discriminant
if POSITIVE, then TWO real solutions
if ZERO, then ONE real solutions
if NEGATIVE, then TWO complex solutions
.
b^2 � 4ac
(-24)^2 � 4(9)(16)
576 - 576 = 0
.
Since discriminate is ZERO --
ONE real solution

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