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

The Correct Option is B

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 \(a=3.\)