Jawab ya lagi mepet

Jawab ya lagi mepet

Jawab:

$1a. x^2-7x+12=0$

$1b. 2x^2+9x+10=0$

$2. x^2+6x+8=0$

$3a. -3\3b. -10\3c. 29\3d. frac{3}{10}\3e. -frac{29}{10}$

Penjelasan dengan langkah-langkah:

$1a. x_{1}=3,x_{2}=4\x_{1}+x_{2}=3+4=7=-frac{b}{a},a=1,b=-7\x_{1}x_{2}=3(4)=12,c=12\ax^2+bx+c=0\x^2-7x+12=0\\1b. x_{1}=frac{1}{2},x_{2}=-5\x_{1}+x_{2}=frac{1}{2}+(-5)=-frac{9}{2}=-frac{b}{a},a=2,b=9\x_{1}x_{2}=(frac{1}{2})(5)=frac{5}{2},c=5\ax^2+bx+c=0\2x^2+9x+5=0$

$2.x_{1}+x_{2}=-6=-frac{b}{a},a=1,b=6\x_{1}x_{2}=8=frac{c}{a},c=8\ax^2+bx+c=0\x^2+6x+8=0$

$x^2+3x-10=0\x_{1}=p,x_{2}=q\3a. p+q=x_{1}+x{2}=-frac{b}{a}=-frac{3}{1}=-3\3b. pcdot q=x_{1}x_{2}=frac{c}{a}=frac{-10}{1}=-10\3c. p^2+q^2=(p+q)^2-2pq=(-3)^2-2(-10)=9+20=29\3d. frac{1}{p}+frac{1}{q}=frac{p+q}{pq}=frac{-3}{-10}=frac{3}{10}\3e. frac{p}{q}+frac{q}{p}=frac{p^2+q^2}{pq}=frac{29}{-10}=-frac{29}{10}$