If x^2+2px-2p+80 for all real values of x,th | vedclass

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(D) Given the quadratic expression $f(x) = x^2+2px-2p+8 > 0$ for all real values of $x$.For a quadratic $ax^2+bx+c > 0$ to be true for all $x \in R$,t

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$(-4,2)$

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(D) Given the quadratic expression $f(x) = x^2+2px-2p+8 > 0$ for all real values of $x$.For a quadratic $ax^2+bx+c > 0$ to be true for all $x \in R$,the conditions are $a > 0$ and the discriminant $D Here,$a = 1 > 0$,which is satisfied.Now,calculate the discriminant $D = b^2 - 4ac $D = (2p)^2 - 4(1)(-2p+8) $4p^2 + 8p - 32 Divide by $4$:$p^2 + 2p - 8 Factorize the quadratic:$(p+4)(p-2) Using the sign scheme method,the expression is negative between the roots $p = -4$ and $p = 2$.Therefore,the set of all possible values of $p$ is $p \in (-4, 2)$.

If $x^2+2px-2p+8>0$ for all real values of $x$,then the set of all possible values of $p$ is

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