( x + a ) 2 = x 2 + 2 a x + a 2 {\displaystyle (x+a)^{2}=x^{2}+2ax+a^{2}}
x 2 + 2 a x = − a 2 {\displaystyle x^{2}+2ax=-a^{2}}
נשלים את הביטוי x 2 + 2 a x {\displaystyle x^{2}+2ax} לריבוע ונקבל x 2 + 2 a x + a 2 {\displaystyle x^{2}+2ax+a^{2}}
נחסיר (או נחבר) את האיבר החדש שנוסף x 2 + 2 a x + a 2 = − a 2 + a 2 {\displaystyle x^{2}+2ax+a^{2}=-a^{2}+a^{2}}
נפתר מהחזקה ונפתור את התרגיל
x 2 − 4 x + 3 = 0 x 2 − 4 x = − 3 x 2 − 4 x + 4 = − 3 + 4 ( x − 2 ) 2 = 1 x − 2 = ± 1 x = ± 1 + 2 x 1 , 2 = 1 , − 1 ( x − 1 ) ( x + 1 ) {\displaystyle {\begin{aligned}x^{2}-4x+3=0\\x^{2}-4x=-3\\x^{2}-4x+4=-3+4\\(x-2)^{2}=1\\x-2=\pm 1\\x=\pm 1+2\\x_{1,2}=1,-1\\(x-1)(x+1)\end{aligned}}}
A ( x 2 + b a x + c a ) {\displaystyle A(x^{2}+{\frac {b}{a}}x+{\frac {c}{a}})}
( b 2 a ) 2 − ( b 2 a ) 2 {\displaystyle \left({\frac {b}{2a}}\right)^{2}-\left({\frac {b}{2a}}\right)^{2}}
A ( x 2 + b a x + ( b 2 a ) 2 ⏟ ( x + b 2 a ) 2 + c a − ( b 2 a ) 2 ) {\displaystyle A\left(\underbrace {x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}} _{\left(x+{\frac {b}{2a}}\right)^{2}}+{\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right)}
נחזיר את הנעלם a A ( ( x + b 2 a ) 2 + c a − ( b 2 a ) 2 ) {\displaystyle A\left(\left(x+{\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right)}
נכפיל בנעלם a A ( x + b 2 a ) 2 + a ∗ c a − a ∗ ( b 2 a ) 2 {\displaystyle A\left(x+{\frac {b}{2a}}\right)^{2}+{\frac {a*c}{a}}-a*\left({\frac {b}{2a}}\right)^{2}}
נצמצם ונפתח את הנעלם בי ונקבל: A ( x + b 2 a ) 2 − ( b 2 4 a ) + c {\displaystyle A\left(x+{\frac {b}{2a}}\right)^{2}-\left({\frac {b^{2}}{4a}}\right)+c}
x 2 + b a x + c a = 0 x 2 + b a x + b 2 4 a 2 − b 2 4 a 2 + c a = 0 ( x + b 2 a ) 2 = b 2 4 a 2 − c a ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = ± b 2 − 4 a c 2 a − b 2 a x 1 , 2 = − b ± b 2 − 4 a c 2 a {\displaystyle {\begin{aligned}x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0\\x^{2}+{\frac {b}{a}}x+{\frac {b^{2}}{4a^{2}}}-{\frac {b^{2}}{4a^{2}}}+{\frac {c}{a}}=0\\(x+{\frac {b}{2a}})^{2}={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}\\(x+{\frac {b}{2a}})^{2}={\frac {b^{2}-4ac}{4a^{2}}}\\x+{\frac {b}{2a}}=\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}\\x=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}-{\frac {b}{2a}}\\x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\end{aligned}}}