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# CAT 2019 Quant Question with Solution 46

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 $b^{2}+c$ ?

1. 3721
2. 549
3. 427
4. 361

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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