Solve for x using the substitution method for the following complex equations

Question

1. x⁴- 5x²+ 4 = 0
2. |x² - 4| < 5
3. f(x) = 2x + 1, if f(f(x)) = 25

Answer 1

1. x⁴- 5x²+ 4 = 0:

Let y = x^2. Then, x^4 = y^2, x^2 = y. Thus, rewrite the equation as.

y^2 -5y + 4 = 0

(y-4)(y-1)=0

y = 4, y = 1.

since x^2 = y, x = +/- sqrt(y).

when y = 4 =>x = +/- 2

when y = 1 =>x=+/- 1

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1. |x² - 4| < 5

This means that -5 < x²-4 < 5.

Add 4 to all parts:

−1<x²<9.

1. The left part: x²>−1 is always true (since x²≥0 for all real x)
2. The right part: x² < 9 means:

−3<x<3

Final Answer:

−3<x<3

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1. f(x) = 2x + 1, if f(f(x)) = 25.

f(f(x)) = 2f(x) + 1 = 25

then f(x) = (25-1)/2 = 12

=> 2x+1 = 12, x = 11/2

Answer 2

https://youtu.be/YCp7mdcI4P4

See the videa for detailed explanations.

Answer 3

1. x² = 1 or x² = 4; x = ±1 or x = ±2
2. x² - 4 ≥ 0 and x² - 4 < 5;

{x ≤ - 2 or x ≥ 2} ∩ {- 3 < x < 3}; {- 3 < x ≤ - 2} U {2 ≤ x <3}

x² - 4 < 0, 4 - x² < 5; x² > - 1; - 2 < x < 2

So, answer: -3 < x < 3

1. ff(x) = 2(2x + 1) + 1 = 4x + 3 = 25; 4x = 22; x = 11/2 = 5.5