If k in R,then the roots of (x-2)(x-3)=k^2 ar | vedclass

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(A) We have,$(x-2)(x-3) = k^2$,where $k \in R$. Expanding the equation: $x^2 - 5x + 6 - k^2 = 0$. Comparing this with the standard quadratic equation

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real and distinct

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(A) We have,$(x-2)(x-3) = k^2$,where $k \in R$. Expanding the equation: $x^2 - 5x + 6 - k^2 = 0$. Comparing this with the standard quadratic equation $ax^2 + bx + c = 0$,we get $a = 1$,$b = -5$,and $c = 6 - k^2$. The discriminant $D$ is given by $D = b^2 - 4ac$. Substituting the values: $D = (-5)^2 - 4(1)(6 - k^2) = 25 - 24 + 4k^2 = 1 + 4k^2$. Since $k^2 \ge 0$ for all $k \in R$,it follows that $1 + 4k^2 \ge 1$. Thus,$D > 0$. Since the discriminant is strictly positive,the roots are always real and distinct.

If $k \in R$,then the roots of $(x-2)(x-3)=k^2$ are always

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