Question

(a+1)x^2+2ax+(a-1)=0

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

(a+1)x^2 + 2ax+(a-1) = 0

Imagine this to be a quadratic equation of the form:

ax^2 + bx + c = 0, Where a is not equal to 0.

Comparing the equations we get,

a = (a+1) ; b = 2a ; c = (a-1)

By quadratic formula, x = b^2 +/-sq.root (b^2 - 4ac)/ 2a

we get x= 4a^2 +/-sq.root [4a^2 - 4{(a+1)(a-1)} ]/ 2(a+1)

x= 4a^2 +/-sq.root [4a^2 - 4(a^2 - b^2)]/2(a+1)

x= 4a^2 +/-sq.root [4a^2 - 4a^2 - 4b^2]/2(a+1)

x= 4a^2 +/- sq.root [4b^2]/2(a+1)

x= 4a^2 +/- 2b / 2(a+1)

x= 2(a^2 +/- b)/ 2(a+1)

x= (a^2+/- b) / a+1

Therefore x =a^2+b/ a+1     (or) a^2-b/ a+1

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