Question

find the value of p for which the quadratic equation (2p+1)x^2-(7p+2)x+(7p-3)=0 has equal roots​

Answer 1

(2p + 1)x² + (7p+2)x + (7p-3) = 0
If they have equal roots,
b²-4ac=0
Let,
a=2p+1
b=7p+2
c=7p-3
Now,
(7p+2)² - 4(2p + 1) (7p-3)=0 
49p² +4+ 28p – 4(14p² +p-3)=0
After simplification,
7p²-24p+16=0 
The roots are,
4 and -4/7.

Hope it helped !

Answer 2

Since the quadratic equation has equal roots, D=0

According to the question

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

If the equation has equal zeros than

$\boxed{ {b}^{2} + 4ac \: = 0}$

Given:

$a = (2p + 1)$

$b = - (7p + 2)$

$c = (7p - 3)$

Putting in the formula

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

$49 {p}^{2} + 4 + 28p - (8p + 4)(7p - 3) = 0$

$\small49 {p}^{2} + 4 + 28p - 56 {p}^{2} + 24p - 28p + 12 = 0$

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

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

Middle term splitting....

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

$7p(p - 4) + 4(p - 4) = 0$

$(7p + 4)(p - 4) = 0$

$p = \frac{ - 4}{7} \: or \: p = 4$