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The quadratic equation $x^{2}+b x+4=0$ will have real roots if
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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
$\alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
$a^{2} \times \frac{c}{a}+a b \times-\frac{b}{a}+b^{2}=a c$
$b^{2} \geq 16$
$b \leq-4, b \geq 4$
$\alpha$ 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
$\alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
$a^{2} \times \frac{c}{a}+a b \times-\frac{b}{a}+b^{2}=a c$
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