[size=150]The expression [math]x^2+8x+16[/math] is equivalent to [math](x+4)^2[/math]. Which expressions are equivalent to [math](x+n)^2[/math] for some number[math]n[/math]?[br][/size][br][math]x^2+10x+25[/math]
Elena and Han solved the equation [math]x^2-6x+7=0[/math] in different ways.[br]Elena said, “First I added 2 to each side:[br][center][math]x^2-6x+7+2=2[/math][/center][br]So that tells me:[br][center][math](x-3)^2=2[/math][/center][br]I can find the square roots of both sides:[br][center][math]x-3=\pm\sqrt{2}[/math][/center][br]Which is the same as:[br][center][math]x=3\pm\sqrt{2}[/math][/center][br]So the two solutions are [math]x=3+\sqrt{2}[/math] and [math]x=3-\sqrt{2}[/math].[br][br]Han said, “I used the quadratic formula:[center][br][math]x=\frac{\text{-}b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}[/math][br][/center]Since [math]x^2-6x+7=0[/math], that means [math]a=1[/math], [math]b=\text{-}6[/math], and [math]c=7[/math]. I know:[br][br][center][math]x=\frac{6\pm\sqrt{36-4\cdot1\cdot7}}{2\cdot1}[/math][/center][br]or[br][center][math]x=\frac{6\pm\sqrt{8}}{2}[/math][/center]So:[br][center][math]x=3\pm\frac{\sqrt{8}}{2}[/math][/center]I think the solutions are [math]x=3+\frac{\sqrt{8}}{2}[/math] and [math]x=3-\frac{\sqrt{8}}{2}[/math]".[br][br]Do you agree with either of them? Explain your reasoning.[br]
[size=150]Under what circumstances would solving an equation of the form [math]x^2+bx+c=0[/math] lead to a solution that doesn’t involve fractions?[/size]
[math]x^2-10x+25=0[/math]
[math]x^2+14x+40=0[/math]
[math]x^2+14x+39=0[/math]
[math]3x^2-5x-11=0[/math]