Question

Solve algebraically:

Answer 1

Answer:

$a)\ x_1=-9,\ x_2=9\\\\b)\ x_1=1,\ x_2=5$

Step-by-step explanation:

Given equations:

$a)\ 2x^2 - 162 = 0\\\\b) -\dfrac{1}{2}(x-3)^2=-2$

A) 2x² - 162 = 0

Step 1: Divide both sides by 2.

$\\\implies \dfrac{2x^2 - 162}{2} = \dfrac{0}{2}\\\\\implies x^2-81=0\\\\\implies x^2=81$

Step 2: Take the square root of both sides (using both the positive and negative roots).

$\\\implies \sqrt{x^2}=\sqrt{81}\\\\\implies x=\pm\ 9$

Step 3: Separate into two cases.

$\implies x_1 = -9,\ x_2 =-9$

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B) -1/2(x - 3)² = -2

Step 1: Multiply both sides by -2.

$\\\implies -2\left(-\dfrac{1}{2}(x-3)^2\right)=-2(-2)\\\\\implies (x-3)^2=4$

Step 2: Take the square root of both sides (using both the positive and negative roots).

$\\\implies\sqrt{(x-3)^2}=\sqrt{4}\\\\\implies x-3=\pm\ 2$

Step 3: Separate into two cases and solve each one.$1)\ x-3=-2\implies x=-2+3\implies \boxed{x=1}\\\\2)\ x-3=2\implies x=2+3\implies \boxed{x=5}$

Answer 2

$\huge\text{Hey there!}$

$\huge\textbf{Equation a.}$

$\rm{2x^2 - 162 = 0}$

$\huge\textbf{Solving for:}$

$\rm{2x^2 - 162 = 0}$

$\huge\textbf{Add \boxed{\bf 162} to both sides:}$

$\rm{2x^2 - 162 + 162 = 0 + 162}$

$\huge\textbf{Simplify it:}$

$\rm{2x^2 = 0 + 162}$

$\rm{2x^2 = 162}$

$\huge\textbf{Divide \boxed{\bf 2} to both sides:}$

$\rm{\dfrac{2x^2}{2} = \dfrac{162}{2}}$

$\huge\textbf{Simplify it:}$

$\rm{x^2 = \dfrac{182}{2}}$

$\rm{x^2 = 81}$

$\huge\textbf{Take the square root of \boxed{\bf 81}}$

$\rm{x = \pm \sqrt{81}}$

$\rm{x = 9\ or\ x = -9}$

$\huge\textbf{Therefore, your answer should be:}$

$\huge\boxed{\textsf{x = } \frak{9\ or } \ \textsf{x = }\frak{-9}}\huge\checkmark$

$\huge\textbf{Equation b.}$

$\rm{-\dfrac{1}{2}(x - 3)^2 = -2}$

$\huge\textbf{Solving for:}$

$\rm{-\dfrac{1}{2}(x - 3)^2 = -2}$

$\huge\textbf{Simplify both sides of your equation:}$

$\rm{-\dfrac{1}{2}x^2 + 3x - \dfrac{9}{2} = -2}$

$\huge\textbf{Subtract \boxed{\bf -2} to both sides:}$

$\rm{-\dfrac{1}{2}x^2 + 3x - \dfrac{9}{2} - (-2) = -2 - (-2)}$

$\huge\textbf{Simplify it:}$

$\rm{- \dfrac{1}{2}x^2 + 3x - \dfrac{5}{2} = 0}$

$\huge\textbf{Now, we can convert the equation to:}$

$\rm{-0.5x^2 + 3x - 2.5 = 0}$

$\large\text{When we changed the equation entirely, we made it easier to solve}\uparrow$

$\huge\textbf{Use the quadratic formula to solve.}$

$\large\textsf{The quadratic formula:}$

$\mathsf{x= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

$\huge\textbf{Here are your labels:}$

$\text{a = }\rm{-0.5}\\\\\text{b = }\rm{ 3}\\\\\rm{c = }\rm{\ -2.5}$

$\huge\textbf{Your new equation:}$

$\rm{x = \dfrac{-(3) \pm \sqrt{3^2 - 4(-0.5)(-2.5)}}{2(-0.5)}}$

$\huge\textbf{Simplify it:}$

$\rm{x = \dfrac{-3 \pm \sqrt{4}}{-1}}$

$\rm{x = 1\ or \ x = 5}$

$\huge\textbf{Therefore, your answer should be: }$

$\huge\boxed{\textsf{x = }\frak{1}\ \textsf{or x = }\frak{5}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~$\frak{Amphitrite1040:)}$