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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

Best Answer 

 #1
avatar+92751 
+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
avatar+92751 
+10
Best Answer

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

 

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

Melody  Aug 23, 2015
 #2
avatar+87293 
0

Very nice, Melody...!!!

 

 

CPhill  Aug 24, 2015
 #3
avatar+92751 
0

Thanks Chris :)))

Melody  Aug 25, 2015

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