Persamaan kuadrat x^2 + 4px +4 =0 mempunyai akar '' x1,dan x2^2 + x1^2x2 = 32 maka nilai p adalah Tolong di bantu kakak semua

PK
[tex]{x}^{2} + 4px + 4 \\ \: = {x1}^{2} + {x2}^{2} \\ ({x1 + x2})^{2} - 2xy = 32 \\ ( - \frac{b}{a} )^{2} - 2( \frac{c}{a}) = 32 \\ ( - \frac{4p}{1})^{2} - 2( \frac{4}{1}) = 32 \\ {16p}^{2} = 40 \\ {p}^{2} = 40 \div 16 \\ p = \sqrt{ \frac{40}{16} } \\ p = \frac{2 \sqrt{10} }{4} \\ p = \frac{1 \sqrt{10} }{2}[/tex]