Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\alpha$ and $\beta(\alpha < \beta)$ be the roots of the equation $x^{2}+b x+c=0$, where $\mathrm{b}>0$ and $\mathrm{c} < 0 .$
Consider the following:
1.$\beta < -\alpha$
2.$\beta < \alpha \mid$
Which of the above is/are correct?
Options:
Consider the following:
1.$\beta < -\alpha$
2.$\beta < \alpha \mid$
Which of the above is/are correct?
Solution:
1798 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
Given quadratic equation, Hence roots of given quadratic equation are
$\beta=\frac{-b+\sqrt{b^{2}-4 c}}{2}$
$\alpha=\frac{-b-\sqrt{b^{2}-4 c}}{2} \quad(\because \alpha < \beta)$
$\Rightarrow-\alpha=\frac{b+\sqrt{b^{2}-4 c}}{2}$ and $\mid \alpha \models \frac{b+\sqrt{b^{2}-4 c}}{2}$
$\beta=\frac{-b+\sqrt{b^{2}-4 c}}{2}$
$\alpha=\frac{-b-\sqrt{b^{2}-4 c}}{2} \quad(\because \alpha < \beta)$
$\Rightarrow-\alpha=\frac{b+\sqrt{b^{2}-4 c}}{2}$ and $\mid \alpha \models \frac{b+\sqrt{b^{2}-4 c}}{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.