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What substitution should be used to rewrite 16(x3 + 1)2 – 22(x3 + 1) – 3 = 0 as a quadratic equation?

User Uzoma
by
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2 Answers

2 votes

Answer:

x^3 +1

Explanation:

User HPP
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1 vote

Answer:

z = x^3 +1

Explanation:

Noting the squared term, it makes sense to substitute for that term:

z = x^3 +1

gives ...

16z^2 -22z -3 = 0 . . . . the quadratic you want

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Solutions derived from that substitution

Factoring gives ...

16z^2 -24z +2z -3 = 0

8z(2z -3) +1(2z -3) = 0

(8z +1)(2z -3) = 0

z = -1/8 or 3/2

Then we can find x:

x^3 +1 = -1/8

x^3 = -9/8 . . . . . subtract 1

x = (-1/2)∛9 . . . . . one of the real solutions

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x^3 +1 = 3/2

x^3 = 1/2 = 4/8 . . . . . . subtract 1

x = (1/2)∛4 . . . . . . the other real solution

The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.

User HamGuy
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