Given the function ƒ(x) = 3x + 5, find ƒ(4) and x such that ƒ(x) = 38.
1 answer:
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Consecutive odd integers are 2 apart
they are x and x+2
the product (x times (x+2)) is 1 less than 4 times their sum
x(x+2)=-1+4(x+x+2)
![x^2+2x=8x+7 minus (8x+7) from both sides [tex]x^2-6x-7=0](https://tex.z-dn.net/?f=x%5E2%2B2x%3D8x%2B7%20minus%20%288x%2B7%29%20from%20both%20sides%20%5Btex%5Dx%5E2-6x-7%3D0)
factor
what numbers multliply to get -7 and add to get -6?
-7 and 1
(x-7)(x+1)=0
set equal to 0
x-7=0
x=7
x+1=0
x=-1
so the x can be 7 or -1
the other number (x+2) can be 7 or 1
test each
7 and 9
7(9)=-1+4(7+9)
63=-1+4(16)
63=-1+64
63=63
true
-1 and 1
-1(1)=-1+4(-1+1)
-1=-1+4(0)
-1=-1+0
-1=-1
true
the 2 integers can be 7 and 9 or -1 and 1 (both pairs of numbers work)
they are x and x+2
the product (x times (x+2)) is 1 less than 4 times their sum
x(x+2)=-1+4(x+x+2)
factor
what numbers multliply to get -7 and add to get -6?
-7 and 1
(x-7)(x+1)=0
set equal to 0
x-7=0
x=7
x+1=0
x=-1
so the x can be 7 or -1
the other number (x+2) can be 7 or 1
test each
7 and 9
7(9)=-1+4(7+9)
63=-1+4(16)
63=-1+64
63=63
true
-1 and 1
-1(1)=-1+4(-1+1)
-1=-1+4(0)
-1=-1+0
-1=-1
true
the 2 integers can be 7 and 9 or -1 and 1 (both pairs of numbers work)
Answer:
3a + 2b
is cube root of 27a^3+54a^2b+36ab^2+8b^3
