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Rebekah Burch
Mathematics
16 October, 02:27
What is the product of 4n^+3=7n?
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Denzel Villanueva
16 October, 02:50
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So if you solve for n
4n^2+3=7n
make into trinomial
subtract 7n from both sides
4n^2-7n+3=0
if we can factor this, then we can asume that the factors are equal to zero becase if
xy=0 then assume x and/or y=0 so
to factor an equation in ax^2+bx+c where a is greater than 1 then
b=t+z
a times c=t times z
to factor 4n^2-7n+3 you do
4 times 3=12
what 2 numbers multiply to get 12 and add to get - 7
the numbers are - 3 and - 4 so
split up the - 7
4n^2-4n-3n+3
group
(4n^2-4n) + (-3n+3)
undistribute
(4n) (n-1) + (-3) (n-1)
reverse distributive property
ab+ac=a (b+c)
(4n-3) (n-1) = 0
set each to zero
4n-3=0
add 3 to both sides
4n=3
divide both sides by 4
n=3/4
n-1=0
add 1 to both sides
n=1
n=3/4 or 1
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