IM Alg1.7.14 Lesson: Completing the Square (Part 3)

[size=150]Elena says, “[math]\left(x+3\right)^2[/math] can be expanded into [math]x^2+6x+9[/math]. Likewise, [math]\left(2x+3\right)^2[/math] can be expanded into [math]4x^2+6x+9[/math].”[/size][br][br]Find an error in Elena’s statement and correct the error. Show your reasoning.
Write each expression in standard form:
[math]\left(4x+1\right)^2[/math]
[math]\left(5x-2\right)^2[/math]
[math]\left(\frac{1}{2}x+7\right)^2[/math]
[math]\left(3x+n\right)^2[/math]
[math]\left(kx+m\right)^2[/math]
Decide if [math]4x^2+12x+9[/math] is a perfect square. If so, write an equivalent expression of the form [math]\left(kx+m\right)^2[/math]. If not, suggest one change to turn it into a perfect square.[br]
Decide if [math]4x^2+8x+25[/math] is a perfect square. If so, write an equivalent expression of the form [math]\left(kx+m\right)^2[/math]. If not, suggest one change to turn it into a perfect square.[br][br]
Find the value of c to make each expression in the left column a perfect square in standard form. Then, write an equivalent expression in the form of squared factors. In the last row, write your own pair of equivalent expressions.
Solve each equation by completing the square:
[math]25x^2+40x=-12[/math]
[math]36x^2-60x+10=-6[/math]
[table][tr][td][size=150]Here are three methods for solving [math]3x^2+8x+5=0[/math]. [br]Try to make sense of each method.[/size][/td][td]Method 1:[/td][/tr][tr][td][/td][td][math]\displaystyle \begin {align}3x^2 + 8x + 5 &= 0\\ (3x + 5)(x + 1) &= 0 \end{align}[/math][br][math]\displaystyle \begin {align} x = \text- \frac53 \quad \text{or} \quad x = \text-1\end {align}[/math][br][/td][/tr][tr][td]Method 2:[/td][td]Method 3:[/td][/tr][tr][td][math]\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\ (3x)^2 + 8(3x) + 15 &= 0\\ U^2 + 8U + 15 &= 0\\ (U+5)(U+3) &= 0 \end{align}[/math][br][math]\displaystyle \begin {align} U = \text-5 \quad &\text{or} \quad U = \text-3\\3x = \text-5 \quad &\text{or} \quad 3x = \text-3\\ x = \text- \frac53 \quad &\text{or} \quad x = \text-1 \end{align}[/math][br][/td][td][math]\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\9x^2 + 24x + 16 &= 1\\(3x + 4)^2 &= 1 \end{align}[/math][br][math]\displaystyle \begin {align} 3x+4 = 1 \quad & \text{or} \quad 3x+4 = \text-1\\x = \text-1 \quad & \text{or} \quad x = \text- \frac53 \end {align}[/math][/td][/tr][/table][br][size=150]Once you understand the methods, use each method at least one time to solve these equations.[/size][br][br][math]5x^2+17x+6=0[/math]
[math]6x^2+19x=-10[/math]
[math]8x^2-33x+4=0[/math]
[math]8x^2-26x=-21[/math]
[math]10x^2+37x=36[/math]
[math]12x^2+20x-77=0[/math]
Find the solutions to [math]3x^2-6x+\frac{9}{4}=0[/math]. Explain your reasoning.
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Information: IM Alg1.7.14 Lesson: Completing the Square (Part 3)