The roots of the equation x2 + (2p-1)x + p2 = 0 are real if

Option d: p ≤ 1/4. Solution: Given the quadratic equation: x2 + (2p-1)x + p2 = 0 Here, the coefficients are: a = 1 , b = 2p-1 , and c = p2 . For the roots of a quadratic equation to be real, the discriminant D must be greater than or equal to zero ( D ≥ 0 ): D = b2 - 4ac ≥ 0 Substitute the coefficients into the discriminant formula: (2p-1)2 - 4(1)(p2) ≥ 0 Expand and simplify the inequality: (4p2 - 4p + 1) - 4p2 ≥ 0 -4p + 1 ≥ 0 1 ≥ 4p 4p ≤ 1 p ≤ 1/4 Therefore, the roots are real if p ≤ 1/4 . Option d

The roots of the equation $x^2 + (2p-1)x + p^2 = 0$ are real if

The correct answer is Option d: $p \le 1/4$. Given the quadratic equation: $$x^2 + (2p-1)x + p^2 = 0$$ Here, the coefficients are: $a = 1$, $b = 2p-1$, and $c = p^2$. For the roots of a quadratic equation to be real, the discriminant $D$ must be greater than or equal to zero ($D \ge 0$): $$D = b^2 - 4ac \ge 0$$ Substitute the coefficients into the discriminant formula: $$(2p-1)^2 - 4(1)(p^2) \ge 0$$ Expand and simplify the inequality: $$(4p^2 - 4p + 1) - 4p^2 \ge 0$$ $$-4p + 1 \ge 0$$ $$1 \ge 4p \implies 4p \le 1 \implies p \le \frac{1}{4}$$ Therefore, the roots are real if $p \le 1/4$. **Option d**

EquationsMCQMTP May 20Question 1073 of 221

All Questions

The roots of the equation $\displaystyle x^2 + (2p-1)x + p^2 = 0$ are real if

Options

\displaystyle p \ge 1

\displaystyle p \le 4

\displaystyle p \ge 1/4

\displaystyle p \le 1/4

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Correct Answer

✅ Option d — $\displaystyle p \le 1/4$

All Options:

A $\displaystyle p \ge 1$
B $\displaystyle p \le 4$
C $\displaystyle p \ge 1/4$
D $\displaystyle p \le 1/4$

Detailed Solution & Explanation

Given the quadratic equation:

x^2 + (2p-1)x + p^2 = 0

Here, the coefficients are:

\displaystyle a = 1

\displaystyle b = 2p-1

, and

\displaystyle c = p^2

.

For the roots of a quadratic equation to be real, the discriminant

\displaystyle D

must be greater than or equal to zero (

\displaystyle D \ge 0

D = b^2 - 4ac \ge 0

Substitute the coefficients into the discriminant formula:

(2p-1)^2 - 4(1)(p^2) \ge 0

Expand and simplify the inequality:

(4p^2 - 4p + 1) - 4p^2 \ge 0

-4p + 1 \ge 0

1 \ge 4p \implies 4p \le 1 \implies p \le \frac{1}{4}

Therefore, the roots are real if

\displaystyle p \le 1/4

.

**Option d**

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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The roots of the equation x2+(2p−1)x+p2=0\displaystyle x^2 + (2p-1)x + p^2 = 0x2+(2p−1)x+p2=0 are real if