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.
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
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