Quadratic eqn

Question

Find the values of p such that 6 lies somewhere between the roots of quadratic equation $x^2 + 2(p-3)x + 9 = 0$ ?

Answer 1

First of all , 

For p to lie between roots of equation the discriminant of given quadratic equation > 0 as roots must exist and they must be distinct..

So ,

          [2(p-3)]²  -  4.9 > 0

==>    4(p-3)² - 4 . 9 > 0

==>    (p-3)²  -  9 > 0

==>     p   >  6  or  p < -6

Also , if we require specifically that 6 lies in between the roots of the equation , and here a = 1 in the quadratic equation ax² + bx + c  =  0 . So as we know in such case the graph is an upward parabola..So

For 6 to lie between the roots , f(6) < 0 is necessary condition ..So substituting x = 6 in the given quadratic polynomial we have :

         6²  +  2.(p-3).6  +  9  <  0

==>   36  + 12 p - 36 + 9  < 0

==>   12 p +  9   < 0

==>    p   <  -3 / 4

So according to discriminant value requirement p > 6 or p < -6 and  here p < -3 / 4 ..

So taking common case we have p < -6 which will satisfy both the properties..

Hence the range of p is : ( -infinity , 6 )

Comments

p < -6 should be correct as discriminant also needs to be satisfied..By the way thanks @vijaycs and @thor for rectifying my silly mistake of not considering p < -6 scenario so getting wrong answer earlier..But now it is correct..

First of all roots must be real and distinct then only there is a meaning of second constraint..So both conditions must be taken into account,

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D > 0 means roots should be real.

D < 0 Roots are not real (complex)

suppose if roots are 6+2i and 6-2i, then also 6 is between these two.

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I think, no need to consider complex roots unless it is explicitly mentioned ..