Question

as raízes da equação 2x2-3px+4q=0 são 1 e 3, sendo assim, determine p e q

Answer 1

$2x^2-3px+4q=0 \\ \\ substituir~~x~~por~~1~~e~~3 \\ \\ 2(1)^2-3p(1)+4q=0 \\ 2-3p+4q=0 \\ \\ -3p+4q=-2~~(I) \\ \\ \\ 2(3)^2-3p(3)+4q=0 \\ 2(9)-9p+4q=0 \\ 18-9p+4q=0 \\ -9p+4q=-18~~~(II) \\ \\ sistema \\ \\ -3p+4p=-2~~(I) \\ -9p+4q=-18~~(II) \\ \\ multiplicar~~(II)~~po~~(-1) \\ \\ -3p+4q=-2 \\ 9p-4q=18 \\ \\ cancelar~~(+4q-4q=0)~~e~~somar \\ \\ -3p+9p=-2+18 \\ 6p=16 \\ \\ p= \frac{16}{6} ~~~simplificando~~~\fbox{ p= \frac{8}{3} }$

$substituir~~p~~em \\ \\ -3p+4q=-2 \\ -\not3.( \frac{8}{\not3} )+4q=-2 \\ -8+4q=-2 \\ 4q=-2+8 \\ 4q=6 \\ q= \frac{6}{4} ~~simplificando~~\fbox{ q= \frac{3}{2} }$

Answer 2

2x²-3px+4q=0 são 1 e 3

P(x)=ax²+bx+c=a * (x-x')*(x-x'')   .....a ≠ 0   e  x' e x'' são as raízes

2x²-3px+4q= = 2 *(x-1)*(x-3)

2x²-3px+4q= = 2 *(x²-3x-x+3)

2x²-3px+4q= = 2 *(x²-4x+3)

2x²-3px+4q= = 2x²-8x+6

-3p=-8  ==> p= 8/3

4q=6  ==> q=6/4=3/2