EquationsMCQPYQ Dec 22Question 1060 of 221
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What will be the value of k\displaystyle k, if the roots of the equation (k4)x22kx+(k+5)=0\displaystyle (k-4)x^2 - 2kx + (k+5) = 0 are equal?

Options

A18
B20
C19
D21
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Correct Answer

Option b20

All Options:

  • A18
  • B20
  • C19
  • D21

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Detailed Solution & Explanation

Given the quadratic equation:
(k4)x22kx+(k+5)=0(k-4)x^2 - 2kx + (k+5) = 0
Comparing with ax2+bx+c=0\displaystyle ax^2 + bx + c = 0, we have:
a=(k4),b=2k,c=(k+5)\displaystyle a = (k-4), b = -2k, c = (k+5)
For the roots to be equal, the discriminant (D\displaystyle D) must be equal to zero (D=0\displaystyle D = 0):
D=b24ac=0D = b^2 - 4ac = 0
Substitute the coefficients into the discriminant formula:
(2k)24(k4)(k+5)=0(-2k)^2 - 4(k-4)(k+5) = 0
4k24(k2+5k4k20)=04k^2 - 4(k^2 + 5k - 4k - 20) = 0
4k24(k2+k20)=04k^2 - 4(k^2 + k - 20) = 0
Divide the entire equation by 4\displaystyle 4:
k2(k2+k20)=0k^2 - (k^2 + k - 20) = 0
k2k2k+20=0k^2 - k^2 - k + 20 = 0
k+20=0    k=20-k + 20 = 0 \implies k = 20
Thus, the value of k\displaystyle k is 20\displaystyle 20, which matches Option b.
**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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