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

Given the quadratic equation \(x^2+bx+c=0\) with roots \(4a\) and \(3a\), we aim to find a possible value for \(b^2+c\). Using Vieta's formulas, the sum and product of the roots of a quadratic equation can be expressed as follows:

\[x_1 + x_2 = -b \quad \text{and} \quad x_1 \cdot x_2 = c\]

Here, \(x_1=4a\) and \(x_2=3a\). Thus we have:

\[4a + 3a = -b \rightarrow 7a = -b\rightarrow b = -7a\]

\[4a \cdot 3a = c \rightarrow 12a^2 = c\]

We now need to calculate \(b^2+c\):

\[b^2+c = (-7a)^2 + 12a^2 = 49a^2 + 12a^2 = 61a^2\]

We need to find the value of \(a\) that makes \(61a^2\) match one of the given options. Let's test each option:

• If \(b^2+c = 549\):
• \[61a^2 = 549\]
• \[a^2 = \frac{549}{61} = 9\]
• \[a = 3\quad(\text{since } a \text{ is integer})\]
• Thus, 549 is a possible value of \(b^2+c\).

Therefore, the correct possible value of \(b^2+c\) is 549.