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Question

Quantitative Aptitude Question on Algebra

If (a+b3)2=52+303(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}, where aa and bb are natural numbers, then a+ba + b equals ?

A

8

B

10

C

9

D

7

Answer

8

Explanation

Solution

We are given the equation (a+b3)2=52+303(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}, where aa and bb are natural numbers.
Expanding the left-hand side:
(a+b3)2=a2+2ab3+3b2(a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2
This gives us two parts: - The rational part: a2+3b2a^2 + 3b^2. - The irrational part: 2ab32ab\sqrt{3}
Equating the rational parts and the irrational parts from both sides of the equation, we get:
1. a2+3b2=521.\ a^2 + 3b^2 = 52, 2. 2ab=302.\ 2ab = 30.
From the second equation, 2ab=302ab = 30, we can solve for abab:
ab=15ab = 15
Now, substitute b=15ab = \frac{15}{a} into the first equation:
a2+3(15a)2=52a^2 + 3\left(\frac{15}{a}\right)^2 = 52
Simplifying:
a2+675a2=52a^2 + \frac{675}{a^2} = 52
Multiply through by a2a^2 to clear the denominator:
a4+675=52a2a^4 + 675 = 52a^2
Rearranging:
a452a2+675=0a^4 - 52a^2 + 675 = 0
Let x=a2x = a^2, so the equation becomes:
x252x+675=0x^2 - 52x + 675 = 0
Solving this quadratic equation using the quadratic formula:
x=52±5224×1×6752×1x = \frac{52 \pm \sqrt{52^2 - 4 \times 1 \times 675}}{2 \times 1}
x=52±270427002x = \frac{52 \pm \sqrt{2704 - 2700}}{2}
x=52±42x = \frac{52 \pm \sqrt{4}}{2}
x=52±22x = \frac{52 \pm 2}{2}
Thus, x=27x = 27 or x=25x = 25. Since x=a2x = a^2, we find that a2=25a^2 = 25, so a=5a = 5.
Now substitute a=5a = 5 into the equation ab=15ab = 15:
5b=15    b=35b = 15 \implies b = 3
Thus, a=5a = 5 and b=3b = 3, so:
a+b=5+3=8a + b = 5 + 3 = 8
Therefore, the correct answer is Option (1).