QUESTION:
Find the value of the discriminant . How will you describe the number and type of roots for 3x² - 6x + 2 = 0?
[tex]\huge\purple{\overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad \ \ \ }}[/tex]
ANSWER:
1 + [tex]\frac {1}{3}[/tex][tex]\sqrt {3}[/tex] or 1 + [tex]\frac {-1}{3}[/tex][tex]\sqrt {3}[/tex]
[tex] \\ [/tex]
SOLUTION:
Step 1: Subtract 2 from both sides.
3x² - 6x + 2 − 2 = 0 − 2
3x2 - 6x = −2
[tex] \\ [/tex]
Step 2: Since the coefficient of 3x^2 is 3, divide both sides by 3.
[tex]\frac {3x²\:-\:6x}\blue{3}[/tex] = [tex]\frac {-2}\blue{3}[/tex]
x² - 2x = [tex]\frac {-2}{3}[/tex]
[tex] \\ [/tex]
Step 3: The coefficient of -2x is -2. Let b=-2.
Then we need to add (b/2)^2=1 to both sides to complete the square.
Add 1 to both sides.
x² − 2x + 1 = [tex]\frac {-2}{3}[/tex] + 1
x² − 2x + 1 = [tex]\frac {1}{3}[/tex]
[tex] \\ [/tex]
Step 4: Factor left side.
(x − 1)² = [tex]\frac {1}{3}[/tex]
[tex] \\ [/tex]
Step 5: Take square root.
x − 1 = ±√[tex]\frac {1}{3}[/tex]
[tex] \\ [/tex]
Step 6: Add 1 to both sides.
x - 1 + 1 = 1 ±√[tex]\frac {1}{3}[/tex]
x = 1 ±[tex]\sqrt\frac {1}{3}[/tex]
The answer is = 1 + [tex]\frac {1}{3}[/tex][tex]\sqrt {3}[/tex] or 1 + [tex]\frac {-1}{3}[/tex][tex]\sqrt {3}[/tex]
[tex]\huge\purple{\overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad \ \ \ }}[/tex]
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