Question

find the value of k for which the quadratic equation (2k+1)x²+2(k+3)x+(k+5)=0 has real and equal roots​

Answer 1

Answer:

k = 52

Step-by-step explanation:

a = 2k+1

b = 2(k+3) = b=2k+6

c = k+5

${b}^{2} - 4ac$

$(2k + 6) {}^{2} - 4 \times (2 k + 1) \times (k + 5)$

$4k {}^{2} + 24k + 36 - 8 {k}^{2} - 20$

$- 4 {k}^{2} + 24 + 16$

$4k {}^{2} - 24k - 16$

$k {}^{2} - 6k - 4$

a = 1

b = -6

c = -4

$b {}^{2} - 4ac$

$( - 6) {}^{2} - 4 \times 1 \times - 4$

36 + 16 = 52

k = 52