CAT 2019 Slot 2QA Question 25

Forming a Quadratic Equation and Relation between roots and coefficientsEasy

The quadratic equation x2 + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b2 + c?

Answer & solution

  • A

    3721

  • 549

  • C

    361

  • D

    427

Solution

Easy

Use Vieta's relations to express bb and cc through aa, form b2+cb^2+c as a single expression in aa, then test which option can equal it for an integer aa.

1

Sum and product of roots. For x2+bx+c=0x^2+bx+c=0 with roots 4a4a and 3a3a:

b=4a+3a=7a(sum of roots)  c=(4a)(3a)=12a2(product of roots)\begin{aligned} &-b = 4a+3a = 7a \quad\text{(sum of roots)}\\ &\ \ c = (4a)(3a) = 12a^2 \quad\text{(product of roots)} \end{aligned}
2

Form b2+cb^2+c. Since b=7ab=-7a, we get b2=49a2b^2=49a^2.

b2+c=49a2+12a2 b2+c=61a2\begin{aligned} &b^2+c = 49a^2 + 12a^2\\ &\Rightarrow\ b^2+c = 61a^2 \end{aligned}
3

Test the options. The value must be 6161 times a perfect square (so aa is an integer).

3721=61×61(61 not a perfect square — reject)361 is not divisible by 61(reject)427=61×7(7 not a perfect square — reject)549=61×9=61×32(a=±3 )\begin{aligned} &3721 = 61\times 61 \quad\text{(}61\text{ not a perfect square — reject)}\\ &361 \ \text{is not divisible by }61 \quad\text{(reject)}\\ &427 = 61\times 7 \quad\text{(}7\text{ not a perfect square — reject)}\\ &549 = 61\times 9 = 61\times 3^2 \quad\text{(}a=\pm 3\ \checkmark\text{)} \end{aligned}
b2+c=549b^2+c = 549
CAT 2019 Slot 2 QA Q25: The quadratic equation x 2 + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following — Solution | TheCATExam