Question

Resuelve las siguientes ecuaciones cuadráticas utilizando el método de completar un trinomio cuadrado perfecto (TCP):

X2+6x-16=0

X2-8x-20=0

3x2-24x+45=0

2x2-11x+15=0

Les agradeceria si me ayudan! :) gracias.

Answer 1

$<var>x^2+6x-16=0\\ x^2+2*3*1x+9-9-16=0\\ x^2+2*3*1x+9=25\\ x^2+2*3*1x+3^2=25\\ (x+3)^2=5^2\\ x+3=5\\ x=2\\ \\ x^2-8x-20=0\\ x^2-2*4*1x+16-16-20=0\\ x^2-2*4*1x+16=16+20\\ x^2-2*4*1x+4^2=36\\ (x-4)^2=6^2\\ x-4=6\\ x=10\\ \\</var>$

$<var>3x^2-24x+45=0 \\ \frac{3x^2-24x+45}{3}=\frac{0}{3}\\ \frac{3x^2}{3}-\frac{24x}{3}+\frac{45}{3}=0\\ x^2-8x+15=0\\ x^2-2*4*1x+16-16+15=0\\ x^2-2*4*1x+16=16-15\\ x^2-2*4*1x+4^2=1\\ (x-4)^2=1^2\\ x-4=1\\ x=5\\ \\ </var>$

$<var>2x^2-11x+15=0\\ \\ \frac{2x^2-11x+15}{2}=\frac{0}{2}\\ \\ \frac{2x^2}{2}-\frac{11x}{2}+\frac{15}{2}=0\\ \\ x^2-\frac{11x}{2}+\frac{15}{2}=0\\ \\ x^2-\frac{2*11*1x}{2*2}+\frac{15}{2}=0\\ \\ x^2-2(\frac{11}{4})(1x)+(\frac{11}{4})^2-(\frac{11}{4})^2 - \frac{15}{2}=0\\ \\ x^2-2(\frac{11}{4})(1x)+(\frac{11}{4})^2=(\frac{11}{4})^2 + \frac{15}{2}\\ \\ (x-\frac{11}{4})^2=\frac{121}{4}+ \frac{15}{2}\\ \\ (x-\frac{11}{4})^2=\frac{121+30}{4}\\ \\ (x-\frac{11}{4})^2=\frac{151}{2^2}\\ \\ </var>$

$<var>(x-\frac{11}{4})^2=\frac{(\sqrt{151})^2}{2^2}\\ \\ (x-\frac{11}{4})^2=(\frac{\sqrt{151}}{2})^2\\ \\ x-\frac{11}{4}=\frac{\sqrt{151}}{2}\\ x=\frac{11}{4}+/-\frac{\sqrt{151}}{2}</var>$