If a-b, a+b are the zeros of x^2-4x+3 then (a, b) = A) (3, 1)
B) (-3,-1)
C) (2, 1)
D) (-2,-1)

α - β , α + β are the zeroes of x² - 4x + 3.

The coordinates of (a,b)

(Let a = α & b = β).

On comparing with standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let ;

a = 1

b = - 4

c = 3.

We know that,Sum of the zeroes = - b/a

⟹ α - β + α + β = - ( - 4)/1

⟹ 2α = 4

⟹ α = 4/2

⟹ α = 2

And,

Product of the zeroes = c/a

⟹ (α - β) * (α + β) = 3/1

(a + b) * (a - b) = a² - b².

⟹ α² - β² = 3

Putting the value of α we get,

⟹ (2)² - β² = 3

⟹ 4 - 3 = β²

⟹ 1 = β²

⟹ √1 = β

⟹ 1 = β

∴(a , b) = (α , β) = (2 , 1). (Option - C).

I hope it will help you.

Regards.

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