If 2x2 - (a+6)x + 12a = 0 , then the roots are:

Option a: 6 and a. Solution: Note: There is a typo in the original question's equation, which should be x2 - (a+6)x + 6a = 0 for the roots to be 6 and a . Let's solve both cases. Case 1: Standard corrected equation x2 - (a+6)x + 6a = 0 x2 - ax - 6x + 6a = 0 x(x - a) - 6(x - a) = 0 (x - 6)(x - a) = 0 This gives the roots as 6 and a , which matches Option a. Case 2: Literal equation 2x2 - (a+6)x + 12a = 0 Using factorization by splitting the middle term where the product of roots is 12a/2 = 6a : If the equation is written as 2x2 - (a+12)x + 6a = 0 , the roots are 6 and a/2 . Under the standard exam interpretation, the roots are 6 and a . Option a

If $2x^2 - (a+6)x + 12a = 0$, then the roots are:

The correct answer is Option a: 6 and $a$. Note: There is a typo in the original question's equation, which should be $x^2 - (a+6)x + 6a = 0$ for the roots to be $6$ and $a$. Let's solve both cases. Case 1: Standard corrected equation $x^2 - (a+6)x + 6a = 0$ $$x^2 - ax - 6x + 6a = 0$$ $$x(x - a) - 6(x - a) = 0 \implies (x - 6)(x - a) = 0$$ This gives the roots as $6$ and $a$, which matches Option a. Case 2: Literal equation $2x^2 - (a+6)x + 12a = 0$ Using factorization by splitting the middle term where the product of roots is $12a/2 = 6a$: If the equation is written as $2x^2 - (a+12)x + 6a = 0$, the roots are $6$ and $a/2$. Under the standard exam interpretation, the roots are $6$ and $a$. **Option a**

EquationsMCQPYQ Nov 20Question 1042 of 221

All Questions

If $\displaystyle 2x^2 - (a+6)x + 12a = 0$ , then the roots are:

Options

A6 and

\displaystyle a

B4 and

\displaystyle a^2

C3 and

\displaystyle 2a

D6 and

\displaystyle 3a

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 6 and $\displaystyle a$

All Options:

A6 and $\displaystyle a$
B4 and $\displaystyle a^2$
C3 and $\displaystyle 2a$
D6 and $\displaystyle 3a$

Detailed Solution & Explanation

Note: There is a typo in the original question's equation, which should be

\displaystyle x^2 - (a+6)x + 6a = 0

for the roots to be

\displaystyle 6

and

\displaystyle a

. Let's solve both cases.
Case 1: Standard corrected equation

\displaystyle x^2 - (a+6)x + 6a = 0

x^2 - ax - 6x + 6a = 0

x(x - a) - 6(x - a) = 0 \implies (x - 6)(x - a) = 0

This gives the roots as

\displaystyle 6

and

\displaystyle a

, which matches Option a.
Case 2: Literal equation

\displaystyle 2x^2 - (a+6)x + 12a = 0

Using factorization by splitting the middle term where the product of roots is

\displaystyle 12a/2 = 6a

:
If the equation is written as

\displaystyle 2x^2 - (a+12)x + 6a = 0

, the roots are

\displaystyle 6

and

\displaystyle a/2

.
Under the standard exam interpretation, the roots are

\displaystyle 6

and

\displaystyle a

.
**Option a**

About This Chapter: Equations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Linear, Quadratic and Cubic Equations

This chapter covers Linear, Quadratic and Cubic Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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If 2x2−(a+6)x+12a=0\displaystyle 2x^2 - (a+6)x + 12a = 02x2−(a+6)x+12a=0, then the roots are: