[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]

[math]x^2+10x+29[/math]

[math]x^2-6x+8[/math]

[math]x^2-6x+9[/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=100[/math]

[math]x^2=38[/math]

[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]

[size=150]What number should be added to the expression [math]x^2-15x[/math] to result in an expression equivalent to a perfect square?[/size]

[size=100]Noah uses the quadratic formula to solve the equation [math]2x^2+3x-5=4[/math]. He finds [math]x=\text{-}2.5[/math] or [math]1[/math]. But, when he checks his answer, he finds that neither -2.5 nor 1 are solutions to the equation. [br]Here are his steps:[/size][br][math]a=2[/math], [math]b=3[/math], [math]c=\text{-}5[/math][br][math]x=\frac{\text{-}3\pm\sqrt{3^2-4\cdot2\cdot\text{-}5}}{2\cdot2}[/math][br][math]x=\frac{\text{-}3\pm\sqrt{49}}{4}[/math][br][math]x=\text{-}2.5[/math] or [math]1[/math][br][br]Explain what Noah’s mistake was.[br]

Solve the equation correctly.[br]

[math]x^2-2x=\text{-}1[/math]

[math]x^2+8x+14=23[/math]

[math]x^2-15=0[/math]

[math]7x^2-2x-5=0[/math]

[math]2x^2+12x=8[/math]

[size=150]What are the solutions to the equation [math]x^2-4x=\text{-}3[/math]?[/size]

[size=150]Which expression is equivalent to [math]\sqrt{\text{-}23}[/math]?[/size]

[size=150]Write each expression in the form [math]a+bi[/math], where [math]a[/math] and [math]b[/math] are real numbers.[/size][br][br][math]5i^2[/math]

[math]i^2\cdot i^2[/math]

[math](\text{-}3i)^2[/math]

[math]7\cdot4i[/math]

[math](5+4i)-(\text{-}3+2i)[/math]

[size=150]Let [math]m=(7-2i)[/math] and [math]k=3i[/math]. Write each expression in the form [math]a+bi[/math], where [math]a[/math] and [math]b[/math] are real numbers.[br][/size][br][math]k-m[/math]

[math]k^2[/math]

[math]m^2[/math]

[math]k\cdot m[/math]