EquationsMCQMTP Dec 22 Series IIQuestion 1092 of 221
All Questions

Find value of x2−10x+1\displaystyle x^2 - 10x + 1, if x=15−26\displaystyle x = 15−26\displaystyle \frac{1}{5 - 2\sqrt{6}}

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — 0\displaystyle 0

All Options:

  • A$25$
  • B1\displaystyle 1
  • C0\displaystyle 0
  • D$49$

Detailed Solution & Explanation

Given the value:
x=15−26x = 15−26\displaystyle \frac{1}{5 - 2\sqrt{6}}
First, let's rationalize the denominator:
x=5+26(5−26)(5+26)=5+2625−(26)2=5+2625−24=5+26x = \displaystyle \frac{5 + 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})} = \frac{5 + 26\displaystyle \sqrt{6}}{25 - (26\displaystyle \sqrt{6})^2} = 5+2625−24=5+2\displaystyle \frac{5 + 2\sqrt{6}}{25 - 24} = 5 + 26\displaystyle \sqrt{6}
Now, isolate the square root term:
x−5=26x - 5 = 26\displaystyle \sqrt{6}
Square both sides of the equation:
(x−5)2=(26)2(x - 5)^2 = (26\displaystyle \sqrt{6})^2
x2−10x+25=24x^2 - 10x + 25 = 24
x2−10x+1=0x^2 - 10x + 1 = 0
Thus, the value of x2−10x+1\displaystyle x^2 - 10x + 1 is exactly 0\displaystyle 0.
This corresponds to Option c.

**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

If 3x+y+2x−y=−1\displaystyle 3x+y+2x−y\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Find value of x2−10x+1\displaystyle x^2 - 10x + 1 if x=15−26\displaystyle x = 15−26\displaystyle \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is $28$. If the cost of an apple is doubled then the cost of $3orangesand\displaystyle oranges and5$ apples is $75$. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of $100questionsof\displaystyle questions of1\displaystyle mark each, a candidate gets $60\% marks. If the candidate attempted all questions and there was a penalty of $0.25$ marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

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