If 2 is a root of \( 2 k^{2}+k t+600 \) find we whe of \( k \) and hence the otber rovt Fer what valuzis) of \( p \) witl the equation \( x^{2}-2 p x-4 p-21-0 \) bave real roors

To find \( k \) when 2 is a root of \( 2k^{2} + kt + 600 \), substitute \( t = 2 \). The equation becomes \( 2k^{2} + 2k + 600 = 0 \). Now, simplify it to \( k^{2} + k + 300 = 0 \). The discriminant (\( b^{2} - 4ac \)) must be non-negative for real roots. Here, it’s \( 1 - 1200 = -1199 \), indicating no real values for \( k \). Next, for the equation \( x^{2} - 2px - 4p - 21 = 0 \) to have real roots, the discriminant must be non-negative: \( (-2p)^{2} - 4(1)(-4p - 21) \geq 0 \). This simplifies to \( 4p^{2} + 16p + 84 \geq 0 \). Factoring out gives us \( (p + 2)(p + 12) \geq 0 \). Hence, \( p \leq -12 \) or \( p \geq -2 \) for real roots.