Question

1. \\(\sqrt{x - 15} = 3 - \sqrt{x}\\)\
2. \\((3x + 7)^{\frac{1}{4}} - 3 = 1\\)

Explanation:

Problem 13: $\boldsymbol{\sqrt{x - 15} = 3 - \sqrt{x}}$

Step1: Isolate and square both sides

To eliminate the square roots, first, we can rearrange the equation to group the square root terms. Let's add $\sqrt{x}$ to both sides:
$\sqrt{x - 15} + \sqrt{x} = 3$
Now, square both sides to get rid of the square roots. Recall that $(a + b)^2 = a^2 + 2ab + b^2$. So:
$(\sqrt{x - 15} + \sqrt{x})^2 = 3^2$
Expanding the left side:
$(x - 15) + 2\sqrt{x(x - 15)} + x = 9$
Simplify the left side:
$2x - 15 + 2\sqrt{x(x - 15)} = 9$

Step2: Isolate the remaining square root

Subtract $2x - 15$ from both sides:
$2\sqrt{x(x - 15)} = 9 - (2x - 15)$
Simplify the right side:
$2\sqrt{x(x - 15)} = 24 - 2x$
Divide both sides by 2:
$\sqrt{x(x - 15)} = 12 - x$

Step3: Square both sides again

Now, square both sides to eliminate the remaining square root:
$(\sqrt{x(x - 15)})^2 = (12 - x)^2$
Simplify both sides:
$x(x - 15) = 144 - 24x + x^2$
Expand the left side:
$x^2 - 15x = 144 - 24x + x^2$

Step4: Solve for $x$

Subtract $x^2$ from both sides:
$-15x = 144 - 24x$
Add $24x$ to both sides:
$9x = 144$
Divide both sides by 9:
$x = 16$

Step5: Check for extraneous solutions

We need to check if $x = 16$ satisfies the original equation because squaring both sides can introduce extraneous solutions.
Left side: $\sqrt{16 - 15} = \sqrt{1} = 1$
Right side: $3 - \sqrt{16} = 3 - 4 = -1$
Since $1
eq -1$, $x = 16$ is an extraneous solution. So, the equation $\sqrt{x - 15} = 3 - \sqrt{x}$ has no solution.

Problem 15: $\boldsymbol{(3x + 7)^{\frac{1}{4}} - 3 = 1}$

Step1: Isolate the radical term

Add 3 to both sides of the equation:
$(3x + 7)^{\frac{1}{4}} = 1 + 3$
Simplify the right side:
$(3x + 7)^{\frac{1}{4}} = 4$

Step2: Eliminate the fourth root

To eliminate the fourth root, raise both sides to the fourth power. Recall that $(a^{\frac{1}{n}})^n = a$ for $a \geq 0$ (since we have a fourth root, the expression inside must be non - negative).
$((3x + 7)^{\frac{1}{4}})^4 = 4^4$
Simplify both sides:
$3x + 7 = 256$

Step3: Solve for $x$

Subtract 7 from both sides:
$3x = 256 - 7$
$3x = 249$
Divide both sides by 3:
$x = \frac{249}{3}=83$

Step4: Check the solution

Substitute $x = 83$ into the original equation:
Left side: $(3(83)+7)^{\frac{1}{4}} - 3=(249 + 7)^{\frac{1}{4}}-3=(256)^{\frac{1}{4}}-3$
Since $256^{\frac{1}{4}}=\sqrt[4]{256} = 4$ (because $4^4=256$), then $4 - 3 = 1$, which is equal to the right side. So, $x = 83$ is a valid solution.

Answer:

s:

• For problem 13: No solution.
• For problem 15: $x = 83$