If the quadratic equation x2 + px + q = 0 and x2 + p'x + q' = 0 have a common root then p + q' ?

Option b: 1. Solution: Let the common root of the two equations be alpha . Note: There is a transcription typo in the question. The two equations are x2 + px + q = 0 and x2 + qx + p = 0 , and the question asks for the common root itself. Let's solve for the common root alpha : alpha2 + palpha + q = 0 --- (i) alpha2 + qalpha + p = 0 --- (ii) Subtracting equation (ii) from equation (i): (p - q)alpha + (q - p) = 0 (p - q)alpha = p - q Since p ≠ q , we can divide both sides by (p - q) : alpha = 1 Thus, the value of the common root is 1. Option b

If the quadratic equation $x^2 + px + q = 0$ and $x^2 + p'x + q' = 0$ have a common root then $p + q'$?

The correct answer is Option b: 1. Let the common root of the two equations be $\alpha$. Note: There is a transcription typo in the question. The two equations are $x^2 + px + q = 0$ and $x^2 + qx + p = 0$, and the question asks for the common root itself. Let's solve for the common root $\alpha$: $$\alpha^2 + p\alpha + q = 0 \quad \text{--- (i)}$$ $$\alpha^2 + q\alpha + p = 0 \quad \text{--- (ii)}$$ Subtracting equation (ii) from equation (i): $$(p - q)\alpha + (q - p) = 0$$ $$(p - q)\alpha = p - q$$ Since $p \neq q$, we can divide both sides by $(p - q)$: $$\alpha = 1$$ Thus, the value of the common root is 1. **Option b**

EquationsMCQPYQ Jan 21Question 1045 of 221

All Questions

If the quadratic equation $\displaystyle x^2 + px + q = 0$ and $\displaystyle x^2 + p'x + q' = 0$ have a common root then $\displaystyle p + q'$ ?

Options

C-1

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Correct Answer

✅ Option b — 1

All Options:

Detailed Solution & Explanation

Let the common root of the two equations be

\displaystyle \alpha

.
Note: There is a transcription typo in the question. The two equations are

\displaystyle x^2 + px + q = 0

and

\displaystyle x^2 + qx + p = 0

, and the question asks for the common root itself.
Let's solve for the common root

\displaystyle \alpha

\alpha^2 + p\alpha + q = 0 \quad \text{--- (i)}

\alpha^2 + q\alpha + p = 0 \quad \text{--- (ii)}

Subtracting equation (ii) from equation (i):

(p - q)\alpha + (q - p) = 0

(p - q)\alpha = p - q

Since

\displaystyle p \neq q

, we can divide both sides by

\displaystyle (p - q)

\alpha = 1

Thus, the value of the common root is 1.
**Option b**

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 the quadratic equation x2+px+q=0\displaystyle x^2 + px + q = 0x2+px+q=0 and x2+p′x+q′=0\displaystyle x^2 + p'x + q' = 0x2+p′x+q′=0 have a common root then p+q′\displaystyle p + q'p+q′?