EquationsMCQPYQ Dec 21Question 1107 of 221
All Questions

Solve x3−7x+6=0\displaystyle x^3 - 7x + 6 = 0

Options

Ax=6,7,−4\displaystyle x = 6, 7, -4
Bx=−1,2,−3\displaystyle x = -1, 2, -3
Cx=1,2,−3\displaystyle x = 1, 2, -3
Dx=2,4,6\displaystyle x = 2, 4, 6
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Correct Answer

✅ Option a — x=6,7,−4\displaystyle x = 6, 7, -4

All Options:

  • Ax=6,7,−4\displaystyle x = 6, 7, -4
  • Bx=−1,2,−3\displaystyle x = -1, 2, -3
  • Cx=1,2,−3\displaystyle x = 1, 2, -3
  • Dx=2,4,6\displaystyle x = 2, 4, 6

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Detailed Solution & Explanation

Let us solve the cubic equation x3−7x+6=0\displaystyle x^3 - 7x + 6 = 0.

By rational root theorem/trial and error, let us test x=1\displaystyle x = 1:
P(1)=13−7(1)+6=1−7+6=0P(1) = 1^3 - 7(1) + 6 = 1 - 7 + 6 = 0
Since P(1)=0\displaystyle P(1) = 0, (x−1)\displaystyle (x-1) is a factor of the polynomial.

Now, we divide x3−7x+6\displaystyle x^3 - 7x + 6 by x−1\displaystyle x - 1 to find the remaining quadratic factor:
x3−7x+6=(x−1)(x2+x−6)x^3 - 7x + 6 = (x-1)(x^2 + x - 6)

Now, factor the quadratic expression x2+x−6\displaystyle x^2 + x - 6:
x2+x−6=(x+3)(x−2)x^2 + x - 6 = (x + 3)(x - 2)

So, the completely factored cubic equation is:
(x−1)(x−2)(x+3)=0(x - 1)(x - 2)(x + 3) = 0

This gives the roots:
x=1,x=2,x=−3\displaystyle x = 1, x = 2, x = -3

Thus, the roots are 1,2,−3\displaystyle 1, 2, -3, which corresponds to Option (c). (Note: The official key marks this as Option (a) due to a typographical error, but mathematically Option (c) is the correct set of roots).

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.

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