Question

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

Answer 1

`-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.!

Answer 2

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)`

Answer 3

`-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`.