NDA Math Mock Test 02 – Question 19

Question:

If \(\alpha\) and \(\beta\) are the roots of the equation \(ax^2 + bx + c = 0\), where \(a \neq 0\), then \((a\alpha + b)(a\beta + b)\) is equal to

यदि \(\alpha\) और \(\beta\) समीकरण \(ax^2 + bx + c = 0\) के मूल हैं, जहाँ \(a \neq 0\) है, तो \((a\alpha + b)(a\beta + b)\) किसके बराबर है?

Options:

  • ab
  • abc
  • ca
  • bc
  • ab
  • abc
  • ca
  • bc

Correct Answer: c)

Explanation:

\((a\alpha + b)(a\beta + b) = a^2\alpha\beta + ab(\alpha+\beta) + b^2\).
Substituting \(\alpha\beta = \dfrac{c}{a}\) and \(\alpha+\beta = -\dfrac{b}{a}\):
\(= a^2\left(\dfrac{c}{a}\right) + ab\left(-\dfrac{b}{a}\right) + b^2 = ac – b^2 + b^2 = ac = ca\).

\((a\alpha + b)(a\beta + b) = a^2\alpha\beta + ab(\alpha+\beta) + b^2\)।
\(\alpha\beta = \dfrac{c}{a}\) तथा \(\alpha+\beta = -\dfrac{b}{a}\) प्रतिस्थापित करने पर:
\(= a^2(c/a) + ab(-b/a) + b^2 = ac – b^2 + b^2 = ac = ca\)।

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top