If the lines represented by the equation 2x^2 | vedclass

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(A) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$. For the lines to be real,the condition is

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$(-\infty, -4] \cup [4, \infty)$

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(A) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$. For the lines to be real,the condition is $h^2 - ab \geq 0$. Comparing $2x^2 - pxy + 2y^2 = 0$ with the general equation,we have $a = 2$,$2h = -p$ (so $h = -p/2$),and $b = 2$. Substituting these values into the condition $h^2 - ab \geq 0$: $(-p/2)^2 - (2)(2) \geq 0$ $p^2/4 - 4 \geq 0$ $p^2 - 16 \geq 0$ $(p - 4)(p + 4) \geq 0$ Solving the inequality,we get $p \leq -4$ or $p \geq 4$. Thus,$p \in (-\infty, -4] \cup [4, \infty)$.

If the lines represented by the equation $2x^2 - pxy + 2y^2 = 0$ are real,then the value of $p$ lies in the interval:

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