Question

Solve each equation using the Quadratic Formula. Watch out for "Special Cases!" x = - 1; x ^ 2 - 5x - 14 = 0; 2x ^ 2 + 3x - 20 = 0; x = 5/4 * 5/2 and x = - 4; x = - 1 and x = - 3; x = 7 7and x = - 2; x ^ 2 + 4x + 3 = 0; x ^ 2 + 2x + 1 = 0 No Real Solution - x ^ 2 ^ 3 = 0

Answer 1

To solve each equation using the quadratic formula, we can use the formula:

x = (-b +/- √(b^2 - 4ac)) / (2a)

Let's solve each equation one by one:

1) x = -1

This equation is already solved for x.

2) x^2 - 5x - 14 = 0

Using the quadratic formula:
a = 1, b = -5, c = -14

x = (-(-5) +/- √((-5)^2 - 4(1)(-14))) / (2(1))
= (5 +/- √(25 + 56)) / 2
= (5 +/- √(81)) / 2
= (5 +/- 9) / 2

x = (5 + 9) / 2 = 14 / 2 = 7
x = (5 - 9) / 2 = -4 / 2 = -2

So the solutions are x = 7 and x = -2.

3) 2x^2 + 3x - 20 = 0

Using the quadratic formula:
a = 2, b = 3, c = -20

x = (-(3) +/- √((3)^2 - 4(2)(-20))) / (2(2))
= (-3 +/- √(9 + 160)) / 4
= (-3 +/- √(169)) / 4
= (-3 +/- 13) / 4

x = (-3 + 13) / 4 = 10 / 4 = 5/2
x = (-3 - 13) / 4 = -16 / 4 = -4

So the solutions are x = 5/2 and x = -4.

4) x = (5/4) * (5/2) and x = -4

This equation is already solved for x.

5) x = -1 and x = -3

This equation is already solved for x.

6) x = 7 and x = -2

This equation is already solved for x.

7) x^2 + 4x + 3 = 0

Using the quadratic formula:
a = 1, b = 4, c = 3

x = (-(4) +/- √((4)^2 - 4(1)(3))) / (2(1))
= (-4 +/- √(16 - 12)) / 2
= (-4 +/- √(4)) / 2
= (-4 +/- 2) / 2

x = (-4 + 2) / 2 = -2 / 2 = -1
x = (-4 - 2) / 2 = -6 / 2 = -3

So the solutions are x = -1 and x = -3.

8) x^2 + 2x + 1 = 0

Using the quadratic formula:
a = 1, b = 2, c = 1

x = (-(2) +/- √((2)^2 - 4(1)(1))) / (2(1))
= (-2 +/- √(4 - 4)) / 2
= (-2 +/- √(0)) / 2
= (-2 +/- 0) / 2

x = (-2 + 0) / 2 = -2 / 2 = -1
x = (-2 - 0) / 2 = -2 / 2 = -1

So the solutions are x = -1 (repeated root).

9) x^2^3 = 0

This equation doesn't follow the form of a quadratic equation, so we cannot solve it using the quadratic formula. There is no real solution to this equation.