Solving a quadratic trinomial equation of the form ax² + bx + c = 0 can be done by first factoring the expression. Secondly, make each bracket equal to 0 and solve each equation. All quadratic trinomials that can be factored will have 1 or 2 solutions. And if the quadratic trinomial cannot be factored then the equation might have no solution or you will have to use the quadratic formula.

Example 1

Solve this equation:

x² + 2x -15 = 0

First you will need to factor the expression.

You need to look for two numbers that multiply to give -15 (the last number) and add to give +2 (the umber before x). These two numbers are -3 and +5 (since -3 x + 5 = -15 and -3 + 5 = 2).

So x² + 2x -15 = 0 can be written as (x-3)(x+5) = 0

Now used the factored expression to make two equations equal to 0 that can be solved:

x – 3 = 0 (add 3 to both sides).

x = 3

x + 5 = 0 (take 5 from both sides).

x = -5

These solutions are sometimes referred to as the roots of the quadratic and are the points where the graph crosses the x axis. See the graph shown above.

Example 2

Solve this equation:

x² -9x + 8 = 0

First you will need to factor the expression.

You need to look for 2 numbers that multiply to give +8 (the last number) and add to give -9 (the umber before x). These two numbers are -1 and -8.

So x² -9x + 8 = 0 can be written as (x-1)(x-8) = 0

Now used the factored expression to make two equations that can be solved:

x – 1 = 0 (add 1 to both sides).

x = 1

x - 8 = 0 (add 8 to both sides).

x = 8

Example 3

2x² = 4 -7x

First you will need to make the trinomial equal to 0 before you can factor the equation. Do this by moving the 4 and -7x to the left hand side of the equation:

2x² + 7x – 4 = 0

Now you should be able to factor the equation. However, this one is harder than the last 2 examples, as the coefficient of x² is not equal to 1. First write down all the factors of -4. These are:

-1 × +4

+1 × -4

+2 × -2

So there are 6 possibilities to choose from for our factored equation. These are:

(2x -1)(x+4)

(2x+4)(x-1)

(2x+1)(x-4)

(2x-4)(x+1)

(2x+2)(x-2)

(2x-2)(x+2)

Now the one that multiplies out to give 2x² + 7x – 4 is the first pair (2x -1)(x+4). Therefore 2x² + 7x – 4 = 0 can be changed to (2x-1)(x+4) = 0

Now used this factored expression to make two equations equal to 0 that you can solve:

2x – 1 = 0 (add 1)

2x = 1 (divide by 2)

x = ½

x + 4 = 0 (take 4)

x = -4

So the two solutions to this quadratic trinomial are x = ½ and x = -4.

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mathscount 3 years ago

i have also done some work on quadratics

http://mathscount.blogspot.in/

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