EquationsMCQMTP Sep 24 Series IQuestion 1101 of 221
All Questions

The equation x2(p+4)x+2p+5=0\displaystyle x^2 - (p + 4)x + 2p + 5 = 0 has equal roots. The value of p\displaystyle p is

Options

A2\displaystyle 2
B2\displaystyle -2
C±2\displaystyle \pm 2
D3\displaystyle 3
For any discrepancies in this question, email contact@cadada.in

Correct Answer

Option c±2\displaystyle \pm 2

All Options:

  • A2\displaystyle 2
  • B2\displaystyle -2
  • C±2\displaystyle \pm 2
  • D3\displaystyle 3

Ad

Detailed Solution & Explanation

For the quadratic equation x2(p+4)x+2p+5=0\displaystyle x^2 - (p + 4)x + 2p + 5 = 0 to have equal (coincident) roots, its discriminant D\displaystyle D must be equal to zero (D=0\displaystyle D = 0).

The given quadratic equation is of the form Ax2+Bx+C=0\displaystyle Ax^2 + Bx + C = 0, where:
A=1\displaystyle A = 1
B=(p+4)\displaystyle B = -(p + 4)
C=2p+5\displaystyle C = 2p + 5

The discriminant D\displaystyle D is:
D=B24ACD = B^2 - 4AC
Setting D=0\displaystyle D = 0:
[(p+4)]24(1)(2p+5)=0[-(p + 4)]^2 - 4(1)(2p + 5) = 0
(p+4)24(2p+5)=0(p + 4)^2 - 4(2p + 5) = 0
p2+8p+168p20=0p^2 + 8p + 16 - 8p - 20 = 0
p24=0p^2 - 4 = 0
p2=4p^2 = 4
p=±2p = \pm 2

Hence, the correct option is **Option (c)**.

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.

More Questions from Equations

Ready to Master Equations?

Practice all 221 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free