SATHEE: Theory of Equations 4 Question 9

Theory of Equations 4 Question 9

10. Let $a, b, c$ be real. If $a x^{2} + b x + c = 0$ has two real roots $α$ and $β$ , where $α < - 1$ and $β > 1$ , then show that $1 + \frac{c}{a} + \frac{b}{a} < 0$

$(1995, 5 M)$

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Solution:

From figure, it is clear that, if $a > 0$ , then $f (- 1) < 0$ and $f (1) < 0$ and if $a < 0, f (- 1) > 0$ and $f (1) > 0$ . In both cases, $a f (- 1) < 0$ and $a f (1) < 0$ .

$\Rightarrow a (a - b + c) < 0 and a (a + b + c) < 0$

On dividing by $a^{2}$ , we get

$1 - \frac{b}{a} + \frac{c}{a} < 0 and 1 + \frac{b}{a} + \frac{c}{a} < 0$

On combining both, we get

$\begin{aligned} 1 \pm \frac{b}{a} + \frac{c}{a} < 0 \\ \Rightarrow & 1 + \frac{b}{a} + \frac{c}{a} < 0 \end{aligned}$

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Theory of Equations 4 Question 9

10. Let a,b,c be real. If ax2+bx+c=0 has two real roots α and β, where α<−1 and β>1, then show that 1+ca+ba<0

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10. Let $a, b, c$ be real. If $a x^{2} + b x + c = 0$ has two real roots $α$ and $β$ , where $α < - 1$ and $β > 1$ , then show that $1 + \frac{c}{a} + \frac{b}{a} < 0$