# Algebra I

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Can anyone summarize the quadratic formula? I am trying to understand this for algebra without any unnecessary and overused words involved like Wikipedia and sites that give me a headache. I just want a paragraph (or more than if its impossible to paraphrase) that is straight, specific and to the point. Thanks in advance.

• Algebra I - ,

Factoring

By far the simplest way of solving quadratic equations is by direct factoring. This does, however, depend on the ability to visualize the exact terms of the factors and definitely improves with experience. Take the following example for instance:
1--Given x^2 + (6/3)x - (35/3) = 0
2--Multiplying through by 3 gives us 3x^2 + 16x - 35 = 0
3--We know that the factors take the form of (ax +/-b)x(cx +/-d)
4--Therefore, we must find values of a, b, c, and d that satisfy ax(x) = 3x^2, ax(+/-d) + cx(+/-b) = +16 and (+/-b)
(+/-d) = -35.
5--Clearly, ether a or c = 3 and b(d) = -5(7) or 5(-7)
6--A little mental arithmetic leads us to a = 3, b = -5, c = 1, and d = +7
7--This leaves us with (3x - 5)(x + 7) = 0
8--If either (3x - 5) or (x + 7) is zero, their product is zero
9--Therefore, 3x - 5 = 0 and x + 7 = 0 making x = +5/3 or -7.

Completing the Square

This method depends on the simplification of the quadratic equation by adding an expression to both sides of the equation that makes one side a perfect square. The process involves the following steps:
1--Simplify and rearrange the equation such that the x^2 and x terms are all on one sides of the equation.
2--Force the coefficient of x^2 to be unity and positive by dividing through by the appropriately signed coeffiecient of x^2.
3--Add the square of half the coefficient of x to both sides of the equation.
4--Take the square root of both sides.
5--Solve the resulting simplified equations.

An example will illustrate the process.

1--Given x^2 - 6x - 16
2--Rearranging, x^2 - 6x = 16
3--Adding (6/2)^2 to both sides gives x^2 - 6x + 9 = 16 + 9 = 25
4--By inspection, x^2 - 6x + 9 = (x - 3)^2 = 25
5--Taking the square root of both sides, (x - 3) = +/-5
6--Therefore, x = 3 + 5 = 8 or x = 3 - 5 = -2.

1--Given 3x^2 = 32 - 10x.
2--Rearranging, 3x^2 + 10x - 32.
3--Dividing through by 3 gives x^2 + (10/3)x = 32/3.
4--Adding (10/3)/2 to both sides gives x^2 + (10/3)x + (5/3)^2 = 32/3 + (5/3)^2 = 121/9.
5--By inspection, x^2 + (10/3)x + (5/3)^2 = (x + (5/3))^2 = 121/9.
6--Taking the square root of both sides, [x + (5/3)] = +/-11/3.
7--Therefore, x = -5/3 + 11/3 = 2 or x = -5/3 - 11/3 = -5 1/3.

Quadratic equations are typically solved by simple factoring or the quadratic formula, x = [-b+/-sqrt(b^2 - 4ac)]/2a. Every quadratic equation can be written in the form ax^2 + bx + c = 0, where a, b, c, may have any numerical values. If we can solve this quadratic, we can solve any quadratic equation.
1--Transposing, we have ax^2 + bx = -c
2--Dividing both sides by a we have x^2 + bx/a = -c/a
3--Adding (b/2a)^2 to each side we have x^2 + bx/a + (b/2a)^2 = b^2/4a^2 - c/a
4--Simplifying we have (x + b/2a)^2 = (b^2 - 4ac)/4a^2
5--Extracting the sqrt we have x + b/2a = +/-[sqrt(b^2 - 4ac)]/2a
6--Therefore, the final quadratic formula becomes
...................x = [-b +/-sqrt(b^2 - 4ac)]
......................................2a

Variations of the quadratic formula are

...................x = -(b/2a) +/-sqrt[(b/2a)^2 - c/a] and

...................x = sqrt[(b/2)^2 - ac] - (b/2)
......................................a

Example: Solve 2X^2 - 38X + 96 = 0.
1--a = 2, b = -38, and c = 96.
2--x = { -(-38) +/- sqrt[(-38^2) - 4(2)(96)]}/2(2)
.......= {38 +/- sqrt[(1444) - 768]}/4
.......= {38 +/- sqrt[676]}/4
.......= {38 +/- 26}/4
.......= {38 + 26}/4 or {38 - 26}/4
.......= 64/4 and 12/4
.....x = 16 and 3.

There is another way to solve quadratics that, in many instances, is just as expedient, if not often simpler. The method requires that the given expression be modified to the form of (mx +/- n)^2 = p giving us solutions of x = (p + n)/m and (p - n)/m. An example will help to visualize the process.

1--Given an expression of the form ax^2 + bx + c = 0 such as x^2 - 10x + 16 = 0.
2--Multiply the expression by a number q that results in q(a) being a perfect square and q(b) being evenly divisible by 2sqrt[q(a)].
3--The number 4 fits our need here resulting in 4x^2 - 40x + 64 = 0.
4--By inspection, we see that 4x^2 - 40x derives from [(sqrt(qa))x - (qb/2sqrt(qa))]^2 or (2x - 10)^2 which results in 4x^2 - 40x + 100.
5--Adding 36 to the right side gives us 4x^2 - 40x + 100 = 36 thereby retaining our original expanded equality.
6--Thus, we end up with (2x - 10)^2 = 36 or (2x - 10) = +6 or -6.
7--Therefore, x = +16/2 = 8 or x = +4/2 = 2.

1--Given 3x^2 - 29x + 154 = 0
2--Multiplying through by 12 yields 36x^2 - 1044x + 5544 = 0
3--By inspection, we see that 36x^2 - 1044x derives from (6x - 87)^2 which gives us 36x^2 - 1044x + 7569.
4--Adding 2025 to the right side gives us 36x^2 - 1044x + 7569 = 2025 thereby retaining our original expanded equality.
5--This leads us to (6x - 87)^2 = 2025 or (6x - 87) = +/-45
6--Then, x = +132/6 = 22 or +42/6 = 7.

Give it a try the next time you are confronted with a quadratic equation to solve. With practice and expreience, it might be just as quick as he quadratic formula.

• Algebra I - ,

7x-y+5