Question

prove that both the roots of the equation (x-a) (x-b) =m^2 are always real and also 10(b) partt pllzzz

Answer 1

$(x - a)(x - b) = \ {m}^{2} \\ {x}^{2} - (a + b)x + ab - {m}^{2} = 0 \\ x = \frac{(a + b) + | \sqrt{ {(a + b) }^{2} - 4(ab - {m}^{2}) } | }{2} \\ x = \frac{(a + b) + | \sqrt{ {(a - b)}^{2} + {m}^{2} } | }{2} \\ here \\ {(a - b)}^{2} + {m}^{2} > 0 \\ > > sum \: of \: squares \: is \: always \: positive < < \\ = > their \: exist \: two \: distinct \: real \: roots \: \\ for \: the \: given \: equation$
hope it helps

Answer 2

We can also prove it by just considering the discriminent(Δ)

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