How do you solve and write the following in interval notation: xx+143(x4)?

1 Answer
Dec 10, 2017

See below

Explanation:

xx+143(x4)

xx+1+43(x4)0

3x(x4)+4(x+1)3(x+1)(x4)0

3x28x+43(x+1)(x4)0

x1,2=b±b24ac2a

Let: a=3; b=-8; c=4

x1,2=8±8243423

x1,2=8±1623

x1,2=8±423

x1=23

x2=2

(x23)(x2)3(x+1)(x4)0

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We need to find when is function 0
function is negative for:
x(1,23][2,4)