Question

If $(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}$, where $a$ and $b$ are natural numbers, then $a + b$ equals ?

• A. 8
• B. 10
• C. 9
• D. 7

Answer 1

The Correct Option is A

We are given the equation $(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}$, where $a$ and $b$ are natural numbers.
Expanding the left-hand side:
\[(a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2\]
This gives us two parts: - The rational part: $a^2 + 3b^2$. - The irrational part: $2ab\sqrt{3}$
Equating the rational parts and the irrational parts from both sides of the equation, we get:
$1.\ a^2 + 3b^2 = 52$, $2.\ 2ab = 30$.
From the second equation, $2ab = 30$, we can solve for $ab$:
\[ab = 15\]
Now, substitute $b = \frac{15}{a}$ into the first equation:
\[a^2 + 3\left(\frac{15}{a}\right)^2 = 52\]
Simplifying:
\[a^2 + \frac{675}{a^2} = 52\]
Multiply through by $a^2$ to clear the denominator:
\[a^4 + 675 = 52a^2\]
Rearranging:
\[a^4 - 52a^2 + 675 = 0\]
Let $x = a^2$, so the equation becomes:
\[x^2 - 52x + 675 = 0\]
Solving this quadratic equation using the quadratic formula:
\[x = \frac{52 \pm \sqrt{52^2 - 4 \times 1 \times 675}}{2 \times 1}\]
\[x = \frac{52 \pm \sqrt{2704 - 2700}}{2}\]
\[x = \frac{52 \pm \sqrt{4}}{2}\]
\[x = \frac{52 \pm 2}{2}\]
Thus, $x = 27$ or $x = 25$. Since $x = a^2$, we find that $a^2 = 25$, so $a = 5$.
Now substitute $a = 5$ into the equation $ab = 15$:
\[5b = 15 \implies b = 3\]
Thus, $a = 5$ and $b = 3$, so:
\[a + b = 5 + 3 = 8\]
Therefore, the correct answer is Option (1).