Question:
The quadratic equation $x^{2}+b x+c=0$ has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of $b^{2}+c$ ?
- 3721
- 549
- 427
- 361
Correct Answer: Option: 2
Sum of roots = 4a+3a=7a=-b
Or b=-7a
Product of roots = 4a×3a =c
Or $c=12{{a}^{2}}$
Now, ${{b}^{2}}+c={{(-7a)}^{2}}+12{{a}^{2}}=61{{a}^{2}}$
Comparing the options.
Option 1: $61{{a}^{2}}=3721$ $\Rightarrow {{a}^{2}}=61$, clearly a is not an integer.
Option 2: $61{{a}^{2}}=549$ $\Rightarrow {{a}^{2}}=9$, we can have a =-3 or 3 (an integer)
Option 3: $61{{a}^{2}}=427$ $\Rightarrow {{a}^{2}}=7$, clearly a is not an integer.
Option 4: $61{{a}^{2}}=361$ $\Rightarrow {{a}^{2}}=\frac{361}{61}$, clearly a is not an integer.
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