If the following equation has no real roots, what are the possible values of p for #x^2 + (3p+1)x - p = 0#?

3 Answers
May 8, 2018

Answer:

# -1 lt p lt -1/9, or, p in (-1,-1/9)#.

Explanation:

We know that, the quadr. eqn.,

#ax^2+bx+c=0" will have no real roots "hArr b^2-4ac lt 0.#

Therefore, in our case, we must have,

#(3p+1)^2-4*1*(-p) lt 0#.

#:. 9p^2+6p+1+4p lt 0, i.e., 9p^2+10p+1 lt 0#.

#:. (p+1)(9p+1) lt 0#.

#:." Case "(1) :(p+1) lt 0 and (9p+1) gt 0, or, #

#" Case "(2) : (p+1) gt 0, and (9p+1) lt 0#.

Case (1) : #(p+1) lt 0 and (9p+1) gt 0#.

#:. p lt -1 and p gt -1/9#. This is impossible.

Similarly, in Case (2) , we have, #p gt -1, &, p lt -1/9, or,#

#-1 lt p lt -1/9#.

We conclude that, # p in (-1, -1/9)#.

Enjoy Maths.!

May 8, 2018

If it has no real roots, then the discriminant of quadratic formula #Delta=(3p+1)^2+4p<0#

#Delta=9p^2+6p++1+4p=9p^2+10p+1<0#

Analyze this last expresion: represents a parabola for all values of p. Lets determine his roots

#p=(-10+-sqrt(100-36))/18=(-10+-8)/18=-1 and -1/9#

So in our case, the values of p for which the first equation has not real roots are in the interval #(-1,-1/9)#

May 8, 2018

Answer:

#-1 < p < -1/9#

Explanation:

If a quadratic equation has no real roots then its discriminant is less than zero. The discriminant of general quadratic equation #ax^2+bx+c=0# is #b^2-4ac#.

We have the equation #x^2+(3p+1)x-p=0# and its discriminant is

#(3p+1)^2-4*1*(-p)=9p^2+6p+1+4p#

= #9p^2+10p+1=(9p+1)(p+1)#

Hence if #x^2+(3p+1)x-p=0# has no real roots then

#(9p+1)(p+1)<0#

Now there are two possibilities

(1) #9p+1<0# and #p+1>0# which means #p<-1/9# and #p> -1#. This is possible when #-1 < p < -1/9#

(2) #9p+1>0# and #p+1<0# which means #p> -1/9# and #p< -1#. But this is just not possible.

Hence for #x^2+(3p+1)x-p=0# to have no real roots

we must have #-1 < p < -1/9#.