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What is the check for x^2-5x+6=0 by completing the square

 Jun 26, 2015

Best Answer 

 #1
avatar+118608 
+5

 

$$\\x^2-5x+6=0\\
x^2-5x=-6\\
x^2-5x+(5/2)^2=-6+(5/2)^2\\
x^2-5x+25/4=-6+25/4\\
(x-5/2)^2=1/4\\\\
x-2.5=\pm 0.5\\
x=2.5\pm 0.5\\
x=3\qquad or \qquad x=2$$

 

check by substitution

$${{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}}$$   true

$${{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}}$$   true                 

 Jun 27, 2015
 #1
avatar+118608 
+5
Best Answer

 

$$\\x^2-5x+6=0\\
x^2-5x=-6\\
x^2-5x+(5/2)^2=-6+(5/2)^2\\
x^2-5x+25/4=-6+25/4\\
(x-5/2)^2=1/4\\\\
x-2.5=\pm 0.5\\
x=2.5\pm 0.5\\
x=3\qquad or \qquad x=2$$

 

check by substitution

$${{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}}$$   true

$${{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}}$$   true                 

Melody Jun 27, 2015

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