For what values of p does the equation (p - 2 | vedclass

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(C) For the quadratic equation $ax^2 + bx + c = 0$ to have no real roots,the discriminant $D = b^2 - 4ac$ must be less than $0$.Here,$a = (p - 2)$,$b

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$p \in (2, 4)$

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(C) For the quadratic equation $ax^2 + bx + c = 0$ to have no real roots,the discriminant $D = b^2 - 4ac$ must be less than $0$.Here,$a = (p - 2)$,$b = 2(p - 2)$,and $c = 2$.If $p = 2$,the equation becomes $0x^2 + 0x + 2 = 0$,which is $2 = 0$,a contradiction. Thus,there are no real roots when $p = 2$.For $p 
eq 2$,$D = [2(p - 2)]^2 - 4(p - 2)(2) $4(p - 2)^2 - 8(p - 2) Divide by $4$: $(p - 2)^2 - 2(p - 2) $(p - 2)(p - 2 - 2) $(p - 2)(p - 4) This inequality holds for $p \in (2, 4)$.Combining the case $p = 2$ (where the equation is not quadratic and has no roots) and the interval $(2, 4)$,we get $p \in [2, 4)$.However,checking the options provided,the most appropriate range covering the condition is $p \in (2, 4)$.

For what values of $p$ does the equation $(p - 2)x^2 + 2(p - 2)x + 2 = 0$ have no real roots?

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