[size=150]In each equation, what expression could be substituted for [math]a[/math] so the equation is true for all values of [math]x[/math]?[/size][br][br][math]x^2=a^2[/math]
[math]\left(3x\right)^2=a^2[/math]
[math]a^2=7x\cdot7x[/math]
[math]a^2=\frac{1}{4}x^2[/math]
[math]a^2=\left(x+1\right)^2[/math]
[math]\left(2x-9\right)\left(2x-9\right)=a^2[/math]
[math]\left(3x\right)^2[/math]
[math]\left(x+1\right)^2[/math]
[math]\left(x-7\right)^2[/math]
[math]\left(x+n\right)^2[/math]
[size=150]Why do you think the following expressions can be described as [b]perfect squares[/b]?[/size][br][br][table][tr][td][math]x^2+6x+9[/math] [/td][td][math]x^2-16x+64[/math] [/td][td][math]x^2+\frac{1}{3}x+\frac{1}{36}[/math] [/td][/tr][/table]
[math]x^4-30x^2+225[/math]
[math]x+14\sqrt{x}+49[/math]
[math]5^{2x}+6\cdot5^x+9[/math]
[table][tr][td][size=150]Han’s method:[/size][br][/td][td][size=150]Jada’s method[/size]:[/td][/tr][tr][td][math]\displaystyle \begin {align} (x-6)^2&=25\\(x-6)(x-6)&=25 \\x^2-12x+36&=25\\ x^2-12x+11&=0\\(x-11)(x-1)&=0\\ \\x=11 \quad \text{or} \quad x&=1 \end{align}[/math][br][/td][td][math]\displaystyle \begin {align} (x-6)^2&=25\\ \\x-6=5 \quad &\text{or} \quad x-6=\text-5\\ x=11 \quad &\text{or} \quad x=1 \end{align}[/math][/td][/tr][/table][br][size=150]Work with a partner to solve these equations. For each equation, one partner solves with Han’s method, and the other partner solves with Jada’s method. Make sure both partners get the same solutions to the same equation. If not, work together to find your mistakes.[/size]