Question

5. If two roots of the quadratic equation (bc)x2+(c-a)x+(a-b)=0 are equal, then let us
prove that, 2b=a+c​

Answer 1

Answer:

that's the solution for this question

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Answer 2

Answer:

Using, the formula

D (Discriminant) of an equation ax²+bx+c is b²-4ac.

Step-by-step explanation:

Because the root os the equation are equal,

D= 0(Discrimant) for roots to be equal

(c-a)² - 4×(a-b)(b-c) = 0

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

=> a²+ (-2b)² + c² + 2×a×c + 2×a×-2b + 2×c ×-2b = 0

Using Identity, a²+b²+c² + 2(ab+bc+ca) = (a+b+c)²

=> ( a - 2b + c)² = 0

=> a + c = 2b

Hence, proved.