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Thread: Quadratic Equation

  1. #1 Quadratic Equation 
    Forum Professor leohopkins's Avatar
    Join Date
    Dec 2006
    Dulwich, London, England
    Hi guys,

    Can someone explain to me how one would go about solving a quadratic equations, in simple laymans terms?

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  3. #2  
    Join Date
    Jan 2008
    you bung the numbers into the standard formula and turn the handle,

    IF A X(Squared) + BX + C = 0 then

    X = -B +/- SQRT(Bsquared - 4AC) /2A

    so if you have 4XSquared + 3X + 7 then

    in the formula above A becomes 4, B=3 and C = 7
    (ie the coefficients of X)

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  4. #3 Re: Quadratic Equation 
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Quote Originally Posted by leohopkins
    Hi guys,

    Can someone explain to me how one would go about solving a quadratic equations, in simple laymans terms?
    The technique is called completing the square and the result is the classic "quadratic formula"


    So, if then which is the classic "quadratic formula".
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  5. #4  
    You can solve quadratics using another, quicker method. It's called factoring, but it can't be done with all quadratic equations, which are in the form , where . Before I get into factoring however, let's consider what a quadratic equation is. It's essentially what's derived through multiplying two binomials (or in the case of a perfect square trinomial, squaring a binomial). Factoring however, isn't always faster than using the quadratic formula, and in some cases not even possible. Let's consider the binomials We can multiply these binomials using the four-step algorithm of FOIL, which is an acronym for First, Inner, Outter, and Last. Namely, we multiply the first terms of our binomials, followed by the inner numbers, than the outter, and finally the last, which is equivalent to the following algebraic expression: , which simplifies to the following quadratic: . It's possible to factor such quadratics, namely those where , but it would probably be more simple to use the quadratic formula.

    Now, let's consider a quadratic that can be factored quickly and easily. When we factor a quadratic where , we must find two numbers which have a sum equal to , and a product equal to . The following quadratic is one such easily solved one: , which derives the binomials . This works since the sum of three and eight is eleven (b), and the product of three and eight is twenty-four (c). What the previous binomials mean is that , and , and insolating x derives the roots (your solution) Something that you should look out for when factoring quadratics is if , such an equation will hold imaginary roots. Also, if , than you're dealing with a squared binomial.

    You can also derive cubic, quartic, or higher order equations through means of raising a binomial to a given power, This can be done by using the binomial theorem (which is an extension of Pascal's triangle). The binomial theorem provides these underlying principles of binomial expansion:

    1. A binomial raised to a given power, will always have terms.

    2. Within the expansion, the first is raised to the power, and decreases by one each term, finally ending the terms with a power of zero, which simply results in one, and can be cancelled out of the equation.

    3. Within the expansion, the first is raised to , and increases by one each term, finally ending at the power.

    4. The sum of the exponents for and within any given terms is equal to .

    5. The coefficient of the term is equal to the following:

    Where is the exponent either variable is raised in the given term.

    It should be noted that factoring quadratics where derives the same roots as when , it's just that the extra steps can be canceled out. If you're interested in learning, write back and I'll explain the method.
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