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Q. The set of values of '`a` ' for which the inequality `x^2-(a+2)x-(a+3)<0` is...

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user8235304 | Student, Grade 11 | Valedictorian

Posted August 6, 2013 at 6:15 PM via web

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Q. The set of values of '`a` ' for which the inequality `x^2-(a+2)x-(a+3)<0` is satisfied for atlleast one positive real `x` is:-?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 6, 2013 at 7:38 PM (Answer #1)

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Find all values of a such that `x^2-(a+2)x-(a+3)<0` is satisfied for at least one positive real x:

We can use the quadratic formula to find the values of x that make the quadratic zero:

`x=((a+2)+-sqrt((a+2)^2-4(1)(-(a+3))))/(2(1))`

`=((a+2)+-sqrt(a^2+8a+16))/2`

`=((a+2)+-|a+4|)/2`

For a+4>0 we get `x=((a+2)+-(a+4))/2=a+3,-1`

For a+4<0 we get `x=((a+2)+-(-a-4))/2=-1,a+3`

Thus the quadratic is zero when x=-1 or x=a+3.

The leading coefficient is positive, so the graph is a parabola opening up. The function will be less than zero between the two zeros.

(1) If a+3<-1 ==> a<-4 the function is less than zero for a+3<x<-1

(2) If a+3=-1 ==> a=-4 the function is never less than zero. It equals zero at x=-1 ( a double root.)

(3) If a+3>-1 the function is less than zero on -1<x<a+3

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The set of all a's such that the function is negative for at least 1 real x is `a!=-4` (Or `(-oo,-4)uu(-4,oo)` , or `a in RR-{-4}` )

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