Question

solve for w, where w is a real number.
\\w - 4 = \sqrt{-5w + 26}\\
(if there is more than one solution, separate them with commas.)

Explanation:

State the equation and constraints

We are given the radical equation:

$$w - 4 = \sqrt{-5w + 26}$$

Since the principal square root is non-negative, we must have:

$$w - 4 \ge 0 \implies w \ge 4$$

Also, the radicand must be non-negative:

$$-5w + 26 \ge 0 \implies w \le \frac{26}{5} = 5.2$$

Thus, any valid solution must satisfy:

$$4 \le w \le 5.2$$

Square both sides of the equation

Square both sides to eliminate the radical:

$$(w - 4)^2 = -5w + 26$$

Expand the left side:

$$w^2 - 8w + 16 = -5w + 26$$

Solve the quadratic equation

Rearrange the terms to form a standard quadratic equation:

$$w^2 - 3w - 10 = 0$$

Factor the quadratic expression:

$$(w - 5)(w + 2) = 0$$

This gives two potential solutions:

$$w = 5 \quad \text{or} \quad w = -2$$

Verify the potential solutions

Check \(w = 5\):

$$\text{Left Side: } 5 - 4 = 1$$

$$\text{Right Side: } \sqrt{-5(5) + 26} = \sqrt{1} = 1$$

Since \(1 = 1\), \(w = 5\) is a valid solution.

Check \(w = -2\):

$$\text{Left Side: } -2 - 4 = -6$$

$$\text{Right Side: } \sqrt{-5(-2) + 26} = \sqrt{36} = 6$$

Since \(-6
e 6\), \(w = -2\) is an extraneous solution.

Thus, the only real solution is \(w = 5\).

Answer:

Solve for \(w\), where \(w\) is a real number.
\(w - 4 = \sqrt{-5w + 26}\)
(If there is more than one solution, separate them with commas.)
\(w =\) <blank>5</blank>