Solve for $$ x $$ : (a) $$ \quad x^{\frac{1}{2}}-3 x^{\frac{1}{4}}+2=0 $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
x^{\frac{1}{2}} - 3x^{\frac{1}{4}} + 2 = 0
$$

**Step 1: Substitution**

Let’s make a substitution to simplify the equation. Let:

$$
y = x^{\frac{1}{4}}
$$

Then:

$$
x^{\frac{1}{2}} = \left(x^{\frac{1}{4}}\right)^2 = y^2
$$

Substituting into the original equation:

$$
y^2 - 3y + 2 = 0
$$

**Step 2: Solve the Quadratic Equation**

The equation $$ y^2 - 3y + 2 = 0 $$ is a quadratic equation in terms of $$ y $$. Solve for $$ y $$:

$$
y = \frac{3 \pm \sqrt{(3)^2 - 4(1)(2)}}{2}
$$
$$
y = \frac{3 \pm \sqrt{9 - 8}}{2}
$$
$$
y = \frac{3 \pm 1}{2}
$$

So, the solutions for $$ y $$ are:

$$
y = 2 \quad \text{or} \quad y = 1
$$

**Step 3: Back-Substitute to Find $$ x $$**

Recall that $$ y = x^{\frac{1}{4}} $$. Therefore:

1. If $$ y = 2 $$:
   $$
   x^{\frac{1}{4}} = 2 \implies x = 2^4 = 16
   $$

2. If $$ y = 1 $$:
   $$
   x^{\frac{1}{4}} = 1 \implies x = 1^4 = 1
   $$

**Conclusion**

The solutions to the equation are:

$$
x = 1 \quad \text{and} \quad x = 16
$$

**Final Answer:**

Both $$ x = 1 $$ and $$ x = 16 $$ satisfy the equation.
The solutions are $$ x = 1 $$ and $$ x = 16 $$.

Solve for \( x \) : (a) \( \quad x^{\frac{1}{2}}-3 x^{\frac{1}{4}}+2=0 \)