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# Solving Quadratic Equations and Substitutions

1. An interesting method for solving quadratic equations came from India.
The steps are: (a) Move the constant term to the right side of the equation (b) Multiply each term in the equation by four times the coefficient of the x2 term (c) Square the coefficient of the original x term and add it to both sides of the equation (d) Take the square root of both sides (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x (f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x
Example: Solve x2 + 3x - 10 = 0
x2 + 3x = 10
4x2 +12x = 40
4x2 + 12x + 9 = 40 + 9
4x2 + 12x + 9 = 49
2x + 3 = ±7
2x + 3 = 7 2x + 3 = 7
2x = 4 2x = -10
x = 2 x = -5
Try these:
(a) x2 - 2x - 13 =0
(b) 4x2 - 4x + 3 = 0
(c) x2 + 12x - 64 = 0
(d) 2x2 - 3x - 5 = 0

2. Mathematicians have been searching for a formula that yields prime numbers. One such formula was x2- x + 41. Select some numbers for x, substitute them in the formula, and see if prime numbers occur. Try to find a number for x that when substituted in the formula yields a composite number.

#### Solution Summary

Step by step solutions to the questions is provided.

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