Matematické Fórum

stuart clark · 30. 10. 2011 17:10

Find the value of $a$ for which $ax^2 + (a - 3)x + 1 < 0$ for at aleast one positive real $x$ .

vanok · 30. 10. 2011 22:04

hi ↑ stuart clark:,
solution of
<<Find the value of $a$ for which $ax^2 + (a - 3)x + 1 < 0$ for at LEAST one positive real x.>>

$ax^2 + (a- 3) x + 1 < 0$ (1)
give for $x=1$ $a + (a- 3) + 1=2a -2 < 0$ what means $a < 1$ (2)

The discriminant of(1) is $D=(a-3)^2-4a=a^2-10a+9=(a-9)(a-1)$
We notice that $D < 0$ for $a \in] 1; 9 [$ and thus (1) does not admit solutions and furthermore it is of constant sign which is positive according to (2)

For $a > 9$ the equation (1) admits 2 roots negatives
$\frac {-a+3 - \sqrt{a^2-10a+9}}{2a}$
$\frac {-a+3 +\sqrt{a^2-10a+9}}{2a}$
because in that case there $\sqrt{a^2-10a+9}-a+3<0$

In more $a =1$ and $a=9$ may be excluded

Conclusion: $a < 1$ verify the conditions of the problem.

Sincerely Vanok

Matematické Fórum

#1 30. 10. 2011 17:10

quadratic inequality

#2 30. 10. 2011 22:04

Re: quadratic inequality

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