Question

The quadratic equation $x^2+bx+c=0$ has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of $b^2+c\ ?$

• A. 3721
• B. 549
• C. 427
• D. 361

Answer 1

Answer: (B)

549

Answer 2

The given quadratic equation has roots whose sum is− b and product is c , where $b=−7a$ and $c=12a^2.$
Now, $b^2+c=(−7a)^2+12a^2=61a^2.$

Comparing this with the given options:
Option 1: $61a^2=3721⇒a^2=61$, clearly a is not an integer.
Option 2: $61a^2=549⇒a^2=9$, which gives $a=−3$ or $a=3.$
Option 3: $61a^2=427⇒a^2=7$, clearly a is not an integer.
Option 4: $61a^2=361⇒a^2=\frac{361}{61}​$, clearly a is not an integer.

Therefore, the only integer solution for a is from Option 2, where $a=−3$ or $a=3.$