Question

prove that the roots of

$(x - a)(x - b) = {h}^{2}$

are always real

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Answer 1

Hello in the above equation coefficient of x i.e b = -(a+b)

Coefficient of x^2 ie. a = 1

Constant term .i.e c = ab - h^2

Discriminant

b^2 - 4 a c

(a+b)^2 - 4ab + h^2 = (a - b)^2 + h^2

So it seems the discriminant is always positive i.e D > 0

Hence only real roots. PROVED

Answer 2

Answer:

Step-by-step explanation:

[x-a] [x-b] =h^2

x^2 -[a+b]x +ab- h^2 =0

A =1

B= -[a+b]

C = ab-h^2

Discriminant

                =   B^2 - 4 AC

= (a+b)^2 - 4ab +4h^2 = (a - b)^2 + 4h^2 ≥ 0

SO   ROOTS ARE REAL