Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The quadratic equation $x^{2}+b x+4=0$ will have real roots if
MathematicsQuadratic EquationNDANDA 2013 (Phase 2)
Options:
  • A $\quad b \leq-4$ only
  • B $b \geq 4$ only
  • C $-4 < b < 4$
  • D $\quad b \leq-4, b \geq 4$
Solution:
2341 Upvotes Verified Answer
The correct answer is: $\quad b \leq-4, b \geq 4$
If root are real $b^{2}-4 \times 4 \geq$
$b^{2} \geq 16$
$b \leq-4, b \geq 4$
$\alpha$ and $\beta$ are roots of given equation. $(a \alpha+b)(a \beta+b)=a^{2} \alpha \beta+a b \alpha+a b \beta+b^{2}$
$=a^{2} \alpha \beta+a b(\alpha+\beta)+b^{2}$
From the given quadratic equation
$\alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
$a^{2} \times \frac{c}{a}+a b \times-\frac{b}{a}+b^{2}=a c$

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.