[tex]\huge\text{Hey there!}[/tex]
[tex]\huge\textbf{Equation a.}[/tex]
[tex]\rm{2x^2 - 162 = 0}[/tex]
[tex]\huge\textbf{Solving for:}[/tex]
[tex]\rm{2x^2 - 162 = 0}[/tex]
[tex]\huge\textbf{Add \boxed{\bf 162} to both sides:}[/tex]
[tex]\rm{2x^2 - 162 + 162 = 0 + 162}[/tex]
[tex]\huge\textbf{Simplify it:}[/tex]
[tex]\rm{2x^2 = 0 + 162}[/tex]
[tex]\rm{2x^2 = 162}[/tex]
[tex]\huge\textbf{Divide \boxed{\bf 2} to both sides:}[/tex]
[tex]\rm{\dfrac{2x^2}{2} = \dfrac{162}{2}}[/tex]
[tex]\huge\textbf{Simplify it:}[/tex]
[tex]\rm{x^2 = \dfrac{182}{2}}[/tex]
[tex]\rm{x^2 = 81}[/tex]
[tex]\huge\textbf{Take the square root of \boxed{\bf 81}}[/tex]
[tex]\rm{x = \pm \sqrt{81}}[/tex]
[tex]\rm{x = 9\ or\ x = -9}[/tex]
[tex]\huge\textbf{Therefore, your answer should be:}[/tex]
[tex]\huge\boxed{\textsf{x = } \frak{9\ or } \ \textsf{x = }\frak{-9}}\huge\checkmark[/tex]
[tex]\huge\textbf{Equation b.}[/tex]
[tex]\rm{-\dfrac{1}{2}(x - 3)^2 = -2}[/tex]
[tex]\huge\textbf{Solving for:}[/tex]
[tex]\rm{-\dfrac{1}{2}(x - 3)^2 = -2}[/tex]
[tex]\huge\textbf{Simplify both sides of your equation:}[/tex]
[tex]\rm{-\dfrac{1}{2}x^2 + 3x - \dfrac{9}{2} = -2}[/tex]
[tex]\huge\textbf{Subtract \boxed{\bf -2} to both sides:}[/tex]
[tex]\rm{-\dfrac{1}{2}x^2 + 3x - \dfrac{9}{2} - (-2) = -2 - (-2)}[/tex]
[tex]\huge\textbf{Simplify it:}[/tex]
[tex]\rm{- \dfrac{1}{2}x^2 + 3x - \dfrac{5}{2} = 0}[/tex]
[tex]\huge\textbf{Now, we can convert the equation to:}[/tex]
[tex]\rm{-0.5x^2 + 3x - 2.5 = 0}[/tex]
[tex]\large\text{When we changed the equation entirely, we made it easier to solve}\uparrow[/tex]
[tex]\huge\textbf{Use the quadratic formula to solve.}[/tex]
[tex]\large\textsf{The quadratic formula:}[/tex]
[tex]\mathsf{x= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}[/tex]
[tex]\huge\textbf{Here are your labels:}[/tex]
[tex]\text{a = }\rm{-0.5}\\\\\text{b = }\rm{ 3}\\\\\rm{c = }\rm{\ -2.5}[/tex]
[tex]\huge\textbf{Your new equation:}[/tex]
[tex]\rm{x = \dfrac{-(3) \pm \sqrt{3^2 - 4(-0.5)(-2.5)}}{2(-0.5)}}[/tex]
[tex]\huge\textbf{Simplify it:}[/tex]
[tex]\rm{x = \dfrac{-3 \pm \sqrt{4}}{-1}}[/tex]
[tex]\rm{x = 1\ or \ x = 5}[/tex]
[tex]\huge\textbf{Therefore, your answer should be: }[/tex]
[tex]\huge\boxed{\textsf{x = }\frak{1}\ \textsf{or x = }\frak{5}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]
[tex]a)\ x_1=-9,\ x_2=9\\\\b)\ x_1=1,\ x_2=5[/tex]
Given equations:
[tex]a)\ 2x^2 - 162 = 0\\\\b) -\dfrac{1}{2}(x-3)^2=-2[/tex]
A) 2x² - 162 = 0
Step 1: Divide both sides by 2.
Step 2: Take the square root of both sides (using both the positive and negative roots).
[tex]\\\implies \sqrt{x^2}=\sqrt{81}\\\\\implies x=\pm\ 9[/tex]
Step 3: Separate into two cases.
[tex]\implies x_1 = -9,\ x_2 =-9[/tex]
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B) -1/2(x - 3)² = -2
Step 1: Multiply both sides by -2.
[tex]\\\implies -2\left(-\dfrac{1}{2}(x-3)^2\right)=-2(-2)\\\\\implies (x-3)^2=4[/tex]
Step 2: Take the square root of both sides (using both the positive and negative roots).
[tex]\\\implies\sqrt{(x-3)^2}=\sqrt{4}\\\\\implies x-3=\pm\ 2[/tex]
Step 3: Separate into two cases and solve each one.[tex]1)\ x-3=-2\implies x=-2+3\implies \boxed{x=1}\\\\2)\ x-3=2\implies x=2+3\implies \boxed{x=5}[/tex]
