Quadratic Equations

Quadratic Equations
I am asked to complete the exercises in the rolls particle on page 397 of Mathematics in Our World. I am to be concise in my reasoning. Please see down the stairs my work for Project iodin and Project Two.
Project One:
Equation (C): x² + 10x - 40 = 0
(a) Move the eternal experimental condition to the right perspective of the par.
X² + 10x - 40 = 0
x² + 10x - 40 + 40 = 0 + 40
x² + 10x + 0 = 0 + 40
x² + 10x = 0 + 40
x² + 10x = 40
(b) Multiply each term in the equation by four times the coefficient of the x unanimousd term.
The co-efficient of the x² term equals 1.
x² + 10x = 40
(4 * 1) * (x² + 10x = 40)
(4) * (x² + 10x = 40)
(4)*x² + (4)*(10x) = (4)*(40)
4x² + 40x = 160
(c) Square the coefficient of the original x term and add it to both sides of the equation.
The co-efficient of the original x term is 10.
(10)² = 100
4x² + 40x = 160
4x² + 40x + 100 = 160+ 100
4x² + 40x + 160 = 260
(d) Take the square root of both sides.
4x² + 40x + 100 = 260
Sqrt(4x² + 40x + 160) = Sqrt(260)
Sqrt(2x + 10)² = Sqrt(10²)
Sqrt(2x + 10)² = ±10
2x + 10 = ±10
(to modify both sides of the equation, divide each side of the equation by 2)
(2x + 10)/2 = ±10/2
x + 6 = ±10
(e) Set the left side of the equation equal to the dogmatic square root of the number on the right side and solve for x.




x + 6 = 10
x + 6 - 6 = 10 - 6
x + 0 = 4
x = 4
(f) Set the left side of equation equal to the negative square root of the number on the right side of the equation and solve for x.
x + 6 = -10
x + 6 - 6 = -10 - 6
x + 0 = -16
x = -16
the final solution is: x ? {-16, 4}
Project Two
Formula that yields prime numbers: x² - x + 41
convey at least five numbers; 0 (zero), devil even, and two odd.
x values selected: 0, 7, 9, 12, 14
if x = 0
x² - x + 41
= 0² - 0 + 41
= 41
if x = 7
x² - x + 41
= 7² - 7 + 41
= 83
if x = 9
x² - x + 41
= 9² - 9 + 41
= 113
if x = 12
x² - x + 41
= 12² - 12 + 41
= 173
if x = 14

x² - x + 41
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