If ` sqrt(x) = sqrt(3) - sqrt(5)` then the value of ` x^(2) - 16 x + 6` is

Options

A. 0
B. `-2`
C. 2
D. 4

Answer

To solve the problem, we start with the equation given:

\[ \sqrt{x} = \sqrt{3} - \sqrt{5} \]

**Step 1: Square both sides to eliminate the square root.**

\[ x = (\sqrt{3} - \sqrt{5})^2 \]

**Hint:** Remember that \((a - b)^2 = a^2 - 2ab + b^2\).

**Step 2: Expand the right side using the formula.**

\[ x = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{5}) + (\sqrt{5})^2 \]

Calculating each term:

\[ x = 3 - 2\sqrt{15} + 5 \]

**Hint:** Calculate the squares of the square roots and the product carefully.

**Step 3: Combine like terms.**

\[ x = 8 - 2\sqrt{15} \]

**Hint:** Combine the constant terms (3 and 5) together.

**Step 4: Now, we need to find \(x^2 - 16x + 6\). First, we calculate \(x^2\).**

\[ x^2 = (8 - 2\sqrt{15})^2 \]

**Hint:** Again use the expansion formula for squaring a binomial.

**Step 5: Expand \(x^2\).**

\[ x^2 = (8)^2 - 2(8)(2\sqrt{15}) + (2\sqrt{15})^2 \]

Calculating each term:

\[ x^2 = 64 - 32\sqrt{15} + 4 \cdot 15 \]

\[ x^2 = 64 - 32\sqrt{15} + 60 \]

**Hint:** Remember that \((2\sqrt{15})^2 = 4 \cdot 15\).

**Step 6: Combine like terms for \(x^2\).**

\[ x^2 = 124 - 32\sqrt{15} \]

**Hint:** Add the constant terms (64 and 60) together.

**Step 7: Now substitute \(x\) and \(x^2\) into the expression \(x^2 - 16x + 6\).**

We already have:

\[ x^2 = 124 - 32\sqrt{15} \]
\[ 16x = 16(8 - 2\sqrt{15}) = 128 - 32\sqrt{15} \]

**Hint:** Multiply \(x\) by 16 carefully.

**Step 8: Substitute into the expression.**

\[ x^2 - 16x + 6 = (124 - 32\sqrt{15}) - (128 - 32\sqrt{15}) + 6 \]

**Step 9: Simplify the expression.**

\[ = 124 - 32\sqrt{15} - 128 + 32\sqrt{15} + 6 \]

The \( -32\sqrt{15} \) and \( +32\sqrt{15} \) cancel out:

\[ = 124 - 128 + 6 \]

**Step 10: Calculate the final result.**

\[ = 2 \]

Thus, the value of \(x^2 - 16x + 6\) is:

\[ \boxed{2} \]