Question

Quadratic 3p^2−7p−1=0

Answer 1

Answer:

$x=\frac{7\pm\sqrt{61} }{6}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Standard Form: ax² + bx + c = 0
• Quadratic Formula: $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Step-by-step explanation:

Step 1: Define

3p² - 7p - 1 = 0

Step 2: Identify Variables

a = 3

b = -7

c = -1

Step 3: Find roots

1. Substitute [Quad Formula]:                   $x=\frac{7\pm\sqrt{(-7)^2-4(3)(-1)} }{2(3)}$
2. Evaluate Exponents:                              $x=\frac{7\pm\sqrt{49-4(3)(-1)} }{2(3)}$
3. Multiply:                                                  $x=\frac{7\pm\sqrt{49+12} }{6}$
4. Add:                                                        $x=\frac{7\pm\sqrt{61} }{6}$

Answer 2

Answer:

x = 7 + √61 ÷ 6

Step-by-step explanation:

3p² - 7p - 1 = 0 is a given equation

3p² - 7p - 1 = 0

Here,

a = 3

b = - 7

c = - 1

Now, Discriminant

D = b² - 4ac

= (- 7)² - 4 (3)(- 1)

= 49 + 12

D = 61 > 0

So, Quadratic Equation

ax² + bx + c = 0

x = - b ± √b² - 4ac ÷ 2a

x = - (- 7) ± √(- 7)² - 4 (3)(- 1) ÷ 2(3)

x = 7 ± √61 ÷ 6

x = 7 ± √61 ÷ 6

x = 7 + √61 ÷ 6 or x = 7 - √61 ÷ 6

∴ Not real Value

Thus, The real value of x is 7 + √61 ÷ 6

-TheUnknownScientist