Question

Find the values of p for which the quadratic equation x2 – 2px + 1 = 0 has
no real roots.
1919
9
Ρ.Τ.Ο​

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

$\begin{lgathered}d = \sqrt{ {b}^{2} - 4ac} \\ d = \sqrt{( - 2p) {}^{2} - 4(1)(1) } \\ d < 0 \\ \sqrt{( - 2p) {}^{2} - 4 } < 0 \\ ( - 2p) {}^{2} < 4 \\ {p}^{2} < 1 \\ p = ( - 1 \: to \: 1)\end{lgathered}d=√b2−4acd=√(−2p)2−4(1)(1)d<0√(−2p)2−4<0(−2p)2<4p2<1p=(−1to1)$

Answer 2

Answer:

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

$$\begin{lgathered}\begin{lgathered}d = \sqrt{ {b}^{2} - 4ac} \\ d = \sqrt{( - 2p) {}^{2} - 4(1)(1) } \\ d < 0 \\ \sqrt{( - 2p) {}^{2} - 4 } < 0 \\ ( - 2p) {}^{2} < 4 \\ {p}^{2} < 1 \\ p = ( - 1 \: to \: 1)\end{lgathered}d=√b2−4acd=√(−2p)2−4(1)(1)d<0√(−2p)2−4<0(−2p)2<4p2<1p=(−1to1)\end{lgathered}$$