Question

if the roots of the quadratic equation a minus b x square + b minus C X + c minus A is equal to zero are equal then prove that 2 is equal to b + c​

Answer 1

Answer:

Step-by-step explanation:

let r & s be the roots, then:

r + s = -(b - c)/(a - b)

but r = s:

2r = -(b - c)/(a - b)

r = -(b - c)/[2(a - b)]

also:

r * s = r^2 = (c - a)/(a - b)

(b - c)^2/[4(a - b)^2] = (c - a)/(a - b)

(b - c)^2/[4(a - b)] = (c - a)

b^2 - 2bc + c^2 = 4ac - 4a^2 - 4bc + 4ab

4a^2 - 4ab + b^2 - 4ac + 2bc + c^2 = 0

(2a - b)^2 - 2c(2a - b) + c^2 = 0

[(2a - b) - c]^2 = 0

2a - b - c = 0

2a = b + c

Answer 2

Hey!!! Liam here is your answer :

(b-c)x2 +(c-a)x +(a-b) = 0

if the roots of this equation in equal than D = o i.e,

D = 0

means..

b2 - 4ac = 0

bere b = (c-a)

a = (b-c)

c = (a-b)

by putting these values in equation 1... we get...

(c-a)2 - 4 (b-c) (a-b) = 0

c2+a2-2ac - 4(ab-b2-ac+bc) = 0

c2+a2-2ac-4ab+4b2+4ac-4bc = 0

a2+4b2+c2-4ab-4bc+2ac = 0

(a-2b+c)2 = 0

so....

a-2b+c = 0

a+c = 2b