Question

Prove that the roots of the equation (x – p)(x - q) = k² are always real.​

Answer 1

Step-by-step explanation:

(x-p)(x-q)=k^2

Without loss of generality let p > q

then take x=|k|+p

we get,(x-p)(x-q)=(|k|) and

(|k|+p-q)

since p-q>0 hence we get

the product(x-p)(x-q)>k^2

Now,note that if x=p then

(x-p)(x-q)=0<k^2

And hence we now have a value of x for which(x-p)(x-q)< k^2

Thus,by intermediate value property

there exist some real number value of x for which (x-p)(x-q)=k^2

Thus the equation

(x-p)(x-q)=k^3=0has at least one real root

Further,we know that complex non-real roots must occur in pairs and hence, the other root cannot be a complex non-real roots.

thus given any k€r

the roots of the equation

(x-p)(x-q)=k^2 must be always be real number

Hence Proved.

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