x^4=(3x-4)^2

To solve the equation \( x^4 = (3x - 4)^2 \), we'll start by expanding the right-hand side:

1. Expand \( (3x - 4)^2 \): \[ (3x - 4)^2 = 9x^2 - 24x + 16 \]

2. Now, rewrite the equation: \[ x^4 = 9x^2 - 24x + 16 \]

3. Move all terms to one side of the equation to set it to zero: \[ x^4 - 9x^2 + 24x - 16 = 0 \]

4. Now we have a polynomial equation in standard form, and let's denote: \[ f(x) = x^4 - 9x^2 + 24x - 16 \]

We can attempt to solve this quartic equation by trying to factor or using numerical methods.

Rational Root Theorem

Let's check for possible rational roots (factors of -16):

Testing \( x = 2 \): \[ f(2) = 2^4 - 9(2^2) + 24(2) - 16 \\ = 16 - 36 + 48 - 16 \\ = 12 \quad (\text{not a root}) \]

Testing \( x = -2 \): \[ f(-2) = (-2)^4 - 9(-2)^2 + 24(-2) - 16 \\ = 16 - 36 - 48 - 16 \\ = -84 \quad (\text{not a root}) \]

Testing \( x = 1 \): \[ f(1) = 1^4 - 9(1^2) + 24(1) - 16 \\ = 1 - 9 + 24 - 16 \\ = 0 \quad (\text{is a root}) \]

Synthetic Division

Since \( x = 1 \) is a root, we can factor \( f(x) \) using synthetic division by \( x - 1 \):

\[ \begin{array}{r|rrrrr} 1 & 1 & 0 & -9 & 24 & -16 \\ & & 1 & 1 & -8 & 16 \\ \hline & 1 & 1 & -8 & 16 & 0 \\ \end{array} \]

This gives us: \[ f(x) = (x - 1)(x^3 + x^2 - 8x + 16) \]

Now, we need to solve the cubic equation \( x^3 + x^2 - 8x + 16 = 0 \).

Finding Roots of the Cubic

We can apply the Rational Root Theorem again to \( x^3 + x^2 - 8x + 16 \).

Testing \( x = 2 \): \[ 2^3 + 2^2 - 8(2) + 16 = 8 + 4 - 16 + 16 = 12 \quad (\text{not a root}) \]

Testing \( x = -2 \): \[ (-2)^3 + (-2)^2 - 8(-2) + 16 = -8 + 4 + 16 + 16 = 28 \quad (\text{not a root}) \]

Testing \( x = -1 \): \[ (-1)^3 + (-1)^2 - 8(-1) + 16 = -1 + 1 + 8 + 16 = 24 \quad (\text{not a root}) \]

Testing \( x = 4 \): \[ 4^3 + 4^2 - 8(4) + 16 = 64 + 16 - 32 + 16 = 64 \quad (\text{not a root}) \]

Finally, testing \( x = -4 \): \[ (-4)^3 + (-4)^2 - 8(-4) + 16 = -64 + 16 + 32 + 16 = 0 \quad (\text{is a root}) \]

More Division

Using synthetic division by \( x + 4 \):

\[ \begin{array}{r|rrrr} -4 & 1 & 1 & -8 & 16 \\ & & -4 & 12 & -16 \\ \hline & 1 & -3 & 4 & 0 \\ \end{array} \]

This gives us: \[ x^3 + x^2 - 8x + 16 = (x + 4)(x^2 - 3x + 4) \]

Solving the Quadratic

Now, solve \( x^2 - 3x + 4 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(4)}}{2(1)} = \frac{3 \pm \sqrt{9 - 16}}{2} \] \[ = \frac{3 \pm \sqrt{-7}}{2} = \frac{3 \pm i\sqrt{7}}{2} \]

Final Solutions

Putting everything together:

1. Real solutions: \( x = 1 \) and \( x = -4 \).
2. Complex solutions: \( x = \frac{3 + i\sqrt{7}}{2} \) and \( x = \frac{3 - i\sqrt{7}}{2} \).

Thus, the solutions to the equation \( x^4 = (3x - 4)^2 \) are: \[ x = 1, \quad x = -4, \quad x = \frac{3 + i\sqrt{7}}{2}, \quad x = \frac{3 - i\sqrt{7}}{2} \]