show your solution need an answer, please need an answer, please.
complete answer pls:(​

Answer:

1. x = -1

2. y = -1, y = 2

3. y = -½ + i [tex]\frac{\sqrt{3} }{2}[/tex]; y = -½ - i [tex]\frac{\sqrt{3} }{2}[/tex]

4. x = -3; x = -6

Step-by-step explanation:

I will only produce work for questions 1 through 4, and you could follow the same steps for questions 5 and 6 so that you could learn and get used to solving quadratic equations. I am practically using the same techniques in solving questions 1 through 4 anyway.

1.) x² + 2x + 1 = 0

where a = 1, b = 2, c = 1

Determine the nature and number of solutions based on the discriminant, b² - 4ac:

b² - 4ac = 2² - 4(1)(1) = 4 - 4 = 0  

This means that the equation has one real root.  

Next, determine the factors of the quadratic equation.  

Use the perfect square trinomial factoring technique:

u² + 2uv + v² = (u + v)²

From the equation, x² + 2x + 1 = 0

where a = 1, b = 2, c = 1

Find factors with product a × c and sum b:

Possible factors:

product a × c :  1 × 1 = 1

sum b :  1 + 1 = 2

Therefore, the binomial factors of x² + 2x + 1 = 0 is (x + 1)²

To find the roots, set x = 0:

x + 1 = 0

Subtract 1 from both sides to isolate x:

x + 1 - 1 = 0 - 1

x = -1  (This is the root of the equation).

2)  y² - y - 2 = 0

where a = 1, b = -1, and c = -2

Determine the nature and number of solutions based on the discriminant, b² - 4ac:  

b²  - 4ac = (-1)² - 4(1)(-2) = 9  

Since b² - 4ac > 0, then it means that the equation will have two real roots.

From the equation, y² - y - 2 = 0

where a = 1, b = -1, and c = -2:

Find factors with product a × c and sum b:

 Product a × c :  

1 × -2 = -2

-1 × 2 = -2

Sum b:  

1+ (-2) = -1

-1 + 2 = 1

Therefore, the possible factors are: 1 and -2:

(y + 1) (y - 2)

To find the roots, set y = 0:

y + 1 = 0

Subtract 1 from both sides:

y + 1 - 1 = 0 - 1

y = -1

y - 2 = 0

Add 2 to both sides:

y -2 + 2 = 0 + 2

y = 2

Therefore, the roots of the quadratic equation, y² - y - 2 = 0 are: y = -1 and y = 2.

3.)  y² + y + 1 = 0

where a = 1, b = 1, and c = 1

Determine the nature and number of solutions based on the discriminant, b² - 4ac:  

b² - 4ac = (1)2 - 4(1)(1) = -3

Since b² - 4ac < 0, then it means that the equation will have two complex roots.

Use the Quadratic Formula:

[tex]y = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}[/tex]

[tex]y = \frac{-1 +/- \sqrt{1^{2} - 4(1)(1)} }{2(1)}[/tex]

[tex]y = \frac{-1 +/- \sqrt{1 - 4} }{2}[/tex]

[tex]y = \frac{-1 +/- \sqrt{-3} }{2}[/tex]

[tex]y = \frac{-1 +/- i\sqrt{3} }{2}[/tex]

Therefore, the roots of the quadratic equation, y² + y + 1 = 0 are:

y = -½ + i [tex]\frac{\sqrt{3} }{2}[/tex]; y = -½ - i [tex]\frac{\sqrt{3} }{2}[/tex]

4) x² + 9x + 18 = 0

where a = 1, b = 9, and c = 18.

Determine the nature and number of solutions based on the discriminant, b² - 4ac:  

b² - 4ac = (9)2 - 4(1)(18) = 9

Since b² - 4ac > 0, then it means that the equation will have two real roots.

Use the Quadratic Formula:

[tex]x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}[/tex]

[tex]x = \frac{-9 +/- \sqrt{9^{2} - 4(1)(18)} }{2(1)}[/tex]

[tex]x = \frac{-9 +/- \sqrt{9} }{2}[/tex]

[tex]x = \frac{-9 + 3}{2}; x = \frac{-9 - 3}{2}[/tex]

[tex]x = \frac{-6}{2}; x = \frac{-12}{2}[/tex]

x = -3; x = -6

Therefore, the roots of the quadratic equation, x² + 9x + 18 = 0

are: x = -3 and x = -6.  

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