Question

find the value if c such that the equation 4x^2 - 2(x + 1)x + ( c+ 4) = 0 has real and equal roots

Answer 1

Answer:

for roots to be equal, determinant D ie b^2-4ac= 0.

therefore putting the values respectively, we will get c = -7/2

Answer 2

Step-by-step explanation:

To find the value of c = ?

Here, A = 4, B = - 2(c + 1) and C = c + 4

∴ D = B^{2} -4ACB

2

−4AC

= (- 2(c + 1))^{2} -4(4)(c + 4)(−2(c+1))

2

−4(4)(c+4)

= (4(c^2+2c+ 1)) -16(c + 4)(4(c

2

+2c+1))−16(c+4)

=4[(c^2+2c+ 1)) -4(c + 4)]=4[(c

2

+2c+1))−4(c+4)]

∵ The roots are real and equal roots,

D ≥ 0

4[(c^2+2c+ 1)) -4(c + 4)]=04[(c

2

+2c+1))−4(c+4)]=0

⇒ c^2+2c+ 1 -4c+16=0c

2

+2c+1−4c+16=0

⇒ c^2-2c-15=0c

2

−2c−15=0

⇒c^2-5c+3c-15=0c

2

−5c+3c−15=0

⇒c(c-5)+3(c-5)=0c(c−5)+3(c−5)=0

⇒ (c+3)(c-5)=0(c+3)(c−5)=0

⇒ (c+3)(c-5)=0

∴ c = 5 or - 3

Hence, the value of "c is 5 or - 3".