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The number of real solutions of the equation  |x-3|-|x+1|/2*|x+1|=1 is ??

Guest Aug 22, 2015

#1
+94203
+10

$$\\|x-3|-|x+1|/2*|x+1|=1\\\\ |x-3|-\frac{|x+1|}{2}*|x+1|=1\\\\ |x-3|-\frac{|x+1|*|x+1|}{2}=1\\\\ |x-3|-\frac{(x+1)^2}{2}=1\\\\ |x-3|=1+\frac{(x+1)^2}{2}\\\\ |x-3|=\frac{x^2+2x+3}{2}\\\\ x-3=\frac{x^2+2x+3}{2}\qquad or \qquad x-3=-\frac{x^2+2x+3}{2} \\\\ 2x-6=x^2+2x+3\qquad or \qquad 2x-6=-x^2-2x-3 \\\\ 0=x^2+9\qquad or \qquad 0=-x^2-4x+3 \\\\ x^2=-9\qquad or \qquad x=\frac{4\pm\sqrt{16+12}}{-2} \\\\$$

$$\\x^2=-9\qquad or \qquad x=\frac{4\pm\sqrt{4*7}}{-2} \\\\ x^2=-9\qquad or \qquad x=\frac{4\pm2\sqrt{7}}{-2} \\\\ x^2=-9\qquad or \qquad x=\frac{2\pm\sqrt{7}}{-1} \\\\ x^2=-9\qquad or \qquad x=-2\pm \sqrt{7 }\\\\ x=\pm\sqrt{- 9}\quad or \quad x=-2+\sqrt{7 } \quad or \quad x=-2-\sqrt{7 }\\\\ So it appears that the only real roots are \\\\ x=-2+\sqrt{7 } \quad or \quad x=-2-\sqrt{7 }\\\\$$

Melody  Aug 23, 2015
#1
+94203
+10

$$\\|x-3|-|x+1|/2*|x+1|=1\\\\ |x-3|-\frac{|x+1|}{2}*|x+1|=1\\\\ |x-3|-\frac{|x+1|*|x+1|}{2}=1\\\\ |x-3|-\frac{(x+1)^2}{2}=1\\\\ |x-3|=1+\frac{(x+1)^2}{2}\\\\ |x-3|=\frac{x^2+2x+3}{2}\\\\ x-3=\frac{x^2+2x+3}{2}\qquad or \qquad x-3=-\frac{x^2+2x+3}{2} \\\\ 2x-6=x^2+2x+3\qquad or \qquad 2x-6=-x^2-2x-3 \\\\ 0=x^2+9\qquad or \qquad 0=-x^2-4x+3 \\\\ x^2=-9\qquad or \qquad x=\frac{4\pm\sqrt{16+12}}{-2} \\\\$$

$$\\x^2=-9\qquad or \qquad x=\frac{4\pm\sqrt{4*7}}{-2} \\\\ x^2=-9\qquad or \qquad x=\frac{4\pm2\sqrt{7}}{-2} \\\\ x^2=-9\qquad or \qquad x=\frac{2\pm\sqrt{7}}{-1} \\\\ x^2=-9\qquad or \qquad x=-2\pm \sqrt{7 }\\\\ x=\pm\sqrt{- 9}\quad or \quad x=-2+\sqrt{7 } \quad or \quad x=-2-\sqrt{7 }\\\\ So it appears that the only real roots are \\\\ x=-2+\sqrt{7 } \quad or \quad x=-2-\sqrt{7 }\\\\$$

Melody  Aug 23, 2015
#2
+93038
0

Very nice, Melody...!!!

CPhill  Aug 24, 2015
#3
+94203
0

Thanks Chris :)))

Melody  Aug 25, 2015