Question

2) determine as raízes de cada equação pelo método de completar quadrado: a) z²+12z-13= 0 b)x² +12x-12= 1 C)x2+13x+40= 0 d) 16x² -16x+2= 34

Answer 1

Equação: $z^2+12z-13 = 0$

Resolvendo a equação pelo método de completar quadrado:
$z^2+12z-13= 0 \\ \\ z^2+12z+36-13= 36 \\ \\ (z+6)^2-13= 36 \\ \\ (z+6)^2= 36+13 \\ \\ (z+6)^2= 49 \\ \\ (z+6)= \pm \sqrt{49} ~~~~ \to ~~~~ z+6= 7 ~~~~ z= 7-6 ~~~~ \boxed{z'= 1}\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~ \to ~~~~ z+6= -7 ~~~~ z= -7-6 ~~~~ \boxed{z''= -13}$

Equação: $x^2+12x-12= 1$

Resolvendo a equação pelo método de completar quadrado:
$x^2+12x-13= 0 \\ \\ x^2+12x+36-13= 36 \\ \\ (x+6)^2-13= 36 \\ \\ (x+6)^2= 36+13 \\ \\ (x+6)^2= 49 \\ \\ (x+6)= \pm \sqrt{49} ~~~~ \to ~~~~ x+6= 7 ~~~~ x= 7-6 ~~~~ \boxed{x'= 1}\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~ \to ~~~~ x+6= -7 ~~~~ x= -7-6 ~~~~ \boxed{x''= -13}$

Equação: $x^2+13x+40= 0$

Resolvendo a equação pelo método de completar quadrado:
$x^2+13x+40= 0 \\ \\ x^2+13x+ \frac{169}{4} +40 = \frac{169}{4} \\ \\ (x+ \frac{13}{2})^2+40= \frac{169}{4} \\ \\ (x+ \frac{13}{2})^2= \frac{169}{4} - 40 \\ \\ (x+ \frac{13}{2})^2= \frac{9}{4} \\ \\ (x+ \frac{13}{2})= \pm \sqrt{ \frac{9}{4} } ~~~~ \to ~~~~ x+ \frac{13}{2} = \frac{3}{2} ~~~~ x= \frac{3}{2}- \frac{13}{2} ~~~~ \boxed{x'= -5}\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~ \to ~~~~ x+ \frac{13}{2} = -\frac{3}{2} ~~~~ x= - \frac{3}{2} - \frac{13}{2} ~~~~ \boxed{x''= -8}$

Equação: $16x^2-16x+2= 34$

Resolvendo a equação pelo método de completar quadrado:
$16x^2-16x-32= 0 \\ \\ x^2-x-2= 0 \\ \\ x^2-x+ \frac{1}{4} -2= \frac{1}{4} \\ \\ (x- \frac{1}{2})^2-2= \frac{1}{4} \\ \\ (x- \frac{1}{2})^2= \frac{1}{4}+ 2 \\ \\ (x- \frac{1}{2})^2= \frac{9}{4} \\ \\ (x- \frac{1}{2})= \pm \sqrt{ \frac{9}{4} } ~~~~ \to ~~~~ x- \frac{1}{2} = \frac{3}{2} ~~~~ x= \frac{3}{2}+ \frac{1}{2} ~~~~ \boxed{x'= 2}\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~ \to ~~~~ x- \frac{1}{2}= - \frac{3}{2} ~~~~ x= - \frac{3}{2}+ \frac{1}{2} ~~~~ \boxed{x''= -1}$