[math]x+x=\frac{1}{4}[/math]
[math]\left(\frac{3}{2}\right)^2=x[/math]
[math]\frac{3}{5}+x=\frac{9}{5}[/math]
[math]\frac{1}{12}+x=\frac{1}{4}[/math]
[math]\left(x-3\right)\left(x+1\right)=5[/math]
[math]x^2+\frac{1}{2}x=\frac{3}{16}[/math]
[math]x^2+3x+\frac{8}{4}=0[/math]
[math]\left(7-x\right)\left(3-x\right)+3=0[/math]
[math]x^2+1.6x+0.63=0[/math]
Show that the equation [math]x^2+10x+9=0[/math] is equivalent to [math]\left(x+3\right)^2+4x=0[/math].[br]
Write an equation that is equivalent to [math]x^2+9x+16=0[/math] and that includes [math]\left(x+4\right)^2[/math].[br]
Does this method help you find solutions to the equations? Explain your reasoning.[br]
[list][size=150][*]Solve one or more of these equations by completing the square.[/*][*]Then, look at the worked solution of the same equation as the one you solved. Find and describe the error or errors in the worked solution.[/*][/size][/list][math]x^2+14x=-24[/math]
Worked solution (with errors):[br][math]\displaystyle \begin {align} x^2 + 14x &= \text-24\\ x^2 + 14x + 28 &= 4\\ (x+7)^2 &= 4\\ \\x+7 = 2 \quad &\text {or} \quad x+7 = \text-2\\ x = \text-5 \quad &\text {or} \quad x = \text-9 \end{align}[/math]
[math]x^2-10x+16=0[/math]
Worked solution (with errors):[br][math]\displaystyle \begin {align} x^2 - 10x + 16 &= 0\\x^2 - 10x + 25 &= 9\\(x - 5)^2 &= 9\\ \\x-5=9 \quad &\text {or} \quad x-5 = \text-9\\ x=14 \quad &\text {or} \quad x=\text-4 \end{align}[/math]
[math]x^2+2.4x=-0.8[/math]
Worked solution (with errors):[br][size=150][size=100][math]\displaystyle \begin {align}x^2 + 2.4x &= \text-0.8\\x^2 + 2.4x + 1.44 &= 0.64\\(x + 1.2)^2&=0.64\\x+1.2 &= 0.8\\ x &=\text -0.4 \end{align}[/math][/size][/size]
[math]x^2-\frac{6}{5}x+\frac{1}{5}=0[/math]
Worked solution (with errors):[br][math]\displaystyle \begin {align} x^2 - \frac65 x + \frac15 &= 0\\x^2 - \frac65 x + \frac{9}{25} &= \frac{9}{25}\\ \left(x-\frac35\right)^2 &= \frac{9}{25}\\ \\x-\frac35= \frac35 \quad &\text {or} \quad x-\frac35=\text- \frac35\\ x=\frac65 \quad &\text {or} \quad x=0 \end{align}[/math]