Question

Find the value of ' p ' for which the quadratic equation x2- 2px + 1 = 0 has no real roots.( Please elaborate aout the confusing case of the roots because i can't understand. )

Answer 1

see there are 3 cases
(1)for real roots d=>0
(2) equal roots d=0
(3) imaginary roots. d<0
here a=1,b=-2p,c=1
$d = \sqrt{ {b}^{2} - 4ac} \\ d = \sqrt{( - 2p) {}^{2} - 4(1)(1) } \\ d < 0 \\ \sqrt{( - 2p) {}^{2} - 4 } < 0 \\ ( - 2p) {}^{2} < 4 \\ {p}^{2} < 1 \\ p = ( - 1 \: to \: 1)$
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Answer 2

Answer: hey friend see there are 3 Roots

(1)for real roots d=>0

(2) equal roots d=0

(3) imaginary roots. d<0

here a=1,b=-2p,c=1

d = \sqrt{ {b}^{2} - 4ac} \\ d = \sqrt{( - 2p) {}^{2} - 4(1)(1) } \\ d < 0 \\ \sqrt{( - 2p) {}^{2} - 4 } < 0 \\ ( - 2p) {}^{2} < 4 \\ {p}^{2} < 1 \\ p = ( - 1 \: to \: 1)

Step-by-step explanation: