Solve the equation. $$ 4(x+1)^{4}-53(x+1)^{2}=-49 $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
4(x+1)^{4} - 53(x+1)^{2} = -49
$$

Let's perform a substitution to simplify the equation. Let:

$$
u = (x + 1)^2
$$

Substituting $$ u $$ into the equation:

$$
4u^2 - 53u = -49
$$

Bring all terms to one side:

$$
4u^2 - 53u + 49 = 0
$$

Now, solve this quadratic equation for $$ u $$:

$$
u = \frac{53 \pm \sqrt{53^2 - 4 \cdot 4 \cdot 49}}{2 \cdot 4}
$$

Calculate the discriminant:

$$
53^2 - 4 \cdot 4 \cdot 49 = 2809 - 784 = 2025 = 45^2
$$

Thus,

$$
u = \frac{53 \pm 45}{8}
$$

This gives two possible values for $$ u $$:

1. $$ u = \frac{53 + 45}{8} = \frac{98}{8} = \frac{49}{4} $$
2. $$ u = \frac{53 - 45}{8} = \frac{8}{8} = 1 $$

Now, revert back to $$ x $$ by solving:

1. $$ (x + 1)^2 = \frac{49}{4} $$

Taking the square root:

$$
   x + 1 = \pm \frac{7}{2}
   $$
   
   So,

$$
   x = -1 + \frac{7}{2} = \frac{5}{2} \quad \text{or} \quad x = -1 - \frac{7}{2} = -\frac{9}{2}
   $$

2. $$ (x + 1)^2 = 1 $$

Taking the square root:

$$
   x + 1 = \pm 1
   $$
   
   So,

$$
   x = 0 \quad \text{or} \quad x = -2
   $$

**Final Solutions:**

$$
x = \frac{5}{2}, \quad x = -\frac{9}{2}, \quad x = 0, \quad x = -2
$$

**Answer:** All real solutions are x = –9⁄2, –2, 0, and 5⁄2.
The solutions to the equation are $$ x = -\frac{9}{2}, -2, 0, $$ and $$ \frac{5}{2} $$.

Solve the equation. \[ 4(x+1)^{4}-53(x+1)^{2}=-49 \]