The values of p for which the equation has real roots is/are ____.(2p + 1)x2-7p+2x+7p-3=0 - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

(2p + 1)

x^{2} - (7 p + 2) x + (7 p - 3) = 0

On comparing the equation with

a x^{2} + bx + c = 0

, we get
a = 2p + 1, b = - (7p + 2) and c = 7p - 3 Since, the roots are equal and real

b^{2} - 4 ac = 0

{(7 p + 2)}^{2} - 4 (2 p + 1) (7 p - 3)

= 0
49

p^{2} + 4 + 28 p - 4 (14 p^{2} - 6 p + 7 p - 3) = 0

p^{2} + 4 + 28 p

- 56

p^{2}

– 4p + 12 = 0
-7

p^{2} + 24 p + 16 = 0

7 p^{2} - 24 p - 16 = 0

{7 p}^{2} - 28 p + 4 p - 16 = 0

7p(p – 4) + 4(p - 4) = 0
(7p + 4) = 0 and (p – 4) = 0
p = -

\frac{4}{7}

and 4

The values of p for which the equation has real roots is/are ____.

(2p + 1) $x^{2} - (7 p + 2) x + (7 p - 3) = 0$