Question:
If the sum of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is equal to the sum of the squares of their reciprocals then \(\dfrac{b^2}{ac} + \dfrac{bc}{a^2}\) is equal to
यदि द्विघात समीकरण \(ax^2 + bx + c = 0\) के मूलों का योग उनके व्युत्क्रमों के वर्गों के योग के बराबर है, तो \(\dfrac{b^2}{ac} + \dfrac{bc}{a^2}\) किसके बराबर है?
Options:
- 1
- -2
- -1
- 2
- 1
- -2
- -1
- 2
Correct Answer: d)
Explanation:
For \(ax^2+bx+c=0\), \(\alpha+\beta = -b/a\) and \(\alpha\beta = c/a\).
Given \(\alpha+\beta = \dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{(\alpha+\beta)^2 – 2\alpha\beta}{(\alpha\beta)^2}\).
\(-\dfrac{b}{a} = \dfrac{b^2/a^2 – 2c/a}{c^2/a^2} = \dfrac{b^2-2ac}{c^2} \implies -bc^2 = ab^2 – 2a^2c \implies ab^2 + bc^2 = 2a^2c\).
Dividing by \(a^2c\): \(\dfrac{b^2}{ac} + \dfrac{bc}{a^2} = 2\).
द्विघात समीकरण के लिए, \(\alpha+\beta = -b/a\) तथा \(\alpha\beta = c/a\)।
प्रश्नानुसार: \(\alpha+\beta = \dfrac{\alpha^2+\beta^2}{\alpha^2\beta^2} = \dfrac{(\alpha+\beta)^2 – 2\alpha\beta}{(\alpha\beta)^2}\)।
\(-\dfrac{b}{a} = \dfrac{b^2-2ac}{c^2} \implies ab^2 + bc^2 = 2a^2c\)।
\(a^2c\) से भाग देने पर: \(\dfrac{b^2}{ac} + \dfrac{bc}{a^2} = 2\)।