Question

find P such that the quadratic equation x^2+5px+16=0 has no real roots.

Answer 1

Given quadratic equation is=x² + 5px +16 = 0

On comparing with standard form of quadratic equation i.e ax² + bx + c =0,

Here, a = 1 , b= 5p, c= 16

D(discriminant)= b²-4ac

= (5p)² - 4× 1 × 16

= 25p² - 64

Since, roots of given equation has no real roots.  D < 0.

25p² - 64 <0

5p² - 8² < 0

(5p +8)(5p -8)< 0                                [ a² - b² = (a-b)(a+b)]

5p +8 <0 or 5p - 8< 0

p > -8/5 or p < 8/5

-8/5 < p  < 8/5

Hence, the required value of p is -8/5 < p  < 8/5.

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

The value p is -8/5 < p <8/5

Step-by-step explanation:

Given the quadratic equation is $x^{2} + 5px + 16$ and it has no real roots.

On comparing with the standard equation $ax^{2} + bx+ c$ we get a = 1,b = 5p,c = 16.

Discriminant D = $b^{2} - 4ac$ < 0

$b^{2} - 4ac$ = $25p^{2} - 4*16 < 0$

which implies -8/5 < p < 8/5