EquationsMCQMTP Nov 21Question 1080 of 221
All Questions

If α,β\displaystyle \alpha, \beta are the roots of equation x2+7x+12=0\displaystyle x^2 + 7x + 12 = 0 then the equation whose roots (α+β)2\displaystyle (\alpha + \beta)^2 and (α−β)2\displaystyle (\alpha - \beta)^2 will be

Options

Ax2−14x+49=0\displaystyle x^2 - 14x + 49 = 0
Bx2−24x+144=0\displaystyle x^2 - 24x + 144 = 0
Cx2−50x+49=0\displaystyle x^2 - 50x + 49 = 0
Dx2−19x+49=0\displaystyle x^2 - 19x + 49 = 0
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — x2−50x+49=0\displaystyle x^2 - 50x + 49 = 0

All Options:

  • Ax2−14x+49=0\displaystyle x^2 - 14x + 49 = 0
  • Bx2−24x+144=0\displaystyle x^2 - 24x + 144 = 0
  • Cx2−50x+49=0\displaystyle x^2 - 50x + 49 = 0
  • Dx2−19x+49=0\displaystyle x^2 - 19x + 49 = 0

Ad

Detailed Solution & Explanation

Given the equation:
x2+7x+12=0x^2 + 7x + 12 = 0
We factor the quadratic expression to find the roots α\displaystyle \alpha and β\displaystyle \beta:
x2+3x+4x+12=0  ⟹  (x+3)(x+4)=0  ⟹  α=−3,β=−4x^2 + 3x + 4x + 12 = 0 \implies (x+3)(x+4) = 0 \implies \alpha = -3, \beta = -4
Now we find the roots of the new equation, which are (α+β)2\displaystyle (\alpha + \beta)^2 and (α−β)2\displaystyle (\alpha - \beta)^2:
1. **First root:**
(α+β)2=(−3−4)2=(−7)2=49(\alpha + \beta)^2 = (-3 - 4)^2 = (-7)^2 = 49
2. **Second root:**
(α−β)2=(−3−(−4))2=(1)2=1(\alpha - \beta)^2 = (-3 - (-4))^2 = (1)^2 = 1
The roots of the new quadratic equation are 49\displaystyle 49 and 1\displaystyle 1.
Calculating the sum (S′\displaystyle S') and product (P′\displaystyle P') of these new roots:
S′=49+1=50S' = 49 + 1 = 50
P′=49×1=49P' = 49 \times 1 = 49
The required quadratic equation is:
x2−S′x+P′=0  ⟹  x2−50x+49=0x^2 - S'x + P' = 0 \implies x^2 - 50x + 49 = 0
This matches 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 \frac{3}{x+y} + \frac{2}{x-y} = -1 and 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 = \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is 28\displaystyle 28. If the cost of an apple is doubled then the cost of 3\displaystyle 3 oranges and 5\displaystyle 5 apples is 75\displaystyle 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 100\displaystyle 100 questions of 1\displaystyle 1 mark each, a candidate gets 60%\displaystyle 60\% marks. If the candidate attempted all questions and there was a penalty of 0.25\displaystyle 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 \frac{3}{x+y} + \frac{2}{x-y} = 3 and 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