What’s up solution should be used to rewrite 16(x^3+1)^2-22(x^3+1)-3=0

Mathematics

1 answer:

ladessa [460]3 years ago

3 0

Answer:

x = cuberoot( -9/8)= -1.04004 or x = cuberoot(1/2)=0.7937

Step-by-step explanation:

let y = x^3 + 1

we have:

16*(y) ^2 - 22 (y) - 3 = 0

16 = 2* 8

-3 = -3* 1

8*-3 + 2 = -24 + 2 = -22

Factor as (8y + 1)(2y - 3) = 16 (y)^2 - 22(y) - 3 = 0

so (8y + 1) = 0 or 2y - 3 = 0

y = -1/8 OR y = 3/2

with y = x^3 + 1

solve for x

-1/8 = x^3 + 1

x^3 = - 9/8

x = cuberoot( -9/8)

about x = -1.04004

or 3/2 = x^3 + 1

1/2 = x^3

x = cuberoot( 1/2 ) = 0.7937

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