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Posts Tagged ‘quadratic’

Quadratic Inequalties

Quadratic inequalities can often cause problems when trying to decide which range of x values are the solutions. As they are quadratics you usually get two values of x and then have to decide whether it is the values of x between the two bounds that are the solutions or the values of x either side.

For example when solving the inequality:
2x2+x-6>0
we get
(2x-3)(x+2)>0

From this you may be tempted to write x<3/2 and x<-2 however this would be incorrect. The solution must be either -2<3/2, or (x<-2 or x>3/2).
This must be the case for any quadratic with two real roots because all the points between the two roots have one sign and these which are outside the roots have another, as can be clearly seen by considering the general graph.

We can determine which of the two solutions to use by either putting in values to see which work or by drawing a sketch, usually drawing a sketch is the best. For the sketch we only need to consider whether the coefficient of x2, in this case 2, is positive or negative so we know which way the graph curves. The y intercept of the graph is irrelevant.

So in our case because 2>=0 the sketch is a “happy” face so the y>0 either side of the roots and therefore our solutions for x are:
x<-2 or x>3/2

By David Woodford

Derive Quadratic Formula

January 28, 2009 13 comments

If you want an explanation on using the formula go to the quadratic formula post (with a downloadable solver)

The quadratic formula,
x=-b +/- sqrt(b*b - 4ac)/2a

for the general equation ax2+bx+c=0

can be derived using the method of “completing the square” as follows.
starting with

ax2+bx+c=0

divide through by a

x2+bx/a+c/a=0

take the c part to the other side

x2+bx/a=-c/a

In order to complete the square we need to add half the coefficient of x (that’s b/2a) squared to both sides so we get

x2+bx/a + b2/4a2=-c/a +b2/4a2

Now we can factorise the left into a squared bracket (you can check this by multiplying it back out)

(x + b/2a)2=-c/a +b2/4a2

So if we square root both sides and take b/2a to the other side we have x on its own

x = -b/2a ±√(-c/a +b2/4a2)

Now to get this in the more traditional form we can take out 1/2a (Which means each term in the root is timesed by 4a2) from the square root to put the whole thing over 2a

x = (-b ±√(b2-4ac))/2a

David Woodford

Factorizing Quadratics

April 16, 2008 1 comment

Factorising quadratics is basically putting them in brackets. In this section we will look at two different ways of factorising quadratics( for simple and complex ones) and when they should be used.

Note/ sometimes a quadratic cannot be factorized using whole numbers, this is when you must use the quadratic equation to find the values of x. See my earlier post and c++ program

Simple type

Use when there is no coefficent of x2

eg)x2+2x-8

start by opening 2 brackets with an x in each
(x )(x )
put the first sign in the first bracket. If the second sign is + put the same sign in both, if its – put the opposite sign in the second bracket
(x+ )(x- )
find the 2 numbers that will add(if both signs in brackets are +) or subtract(if the signs in the brackets are different) to make the middle number(2) and multiply to make the end number(8)
(x+4)(x-2)
and thats your quadratic factorized

Complex Type

Use when there is a coefficent of x2

eg) 8x2-14x-15
Before we can open brackets we need to split up the 14x
8x2 ?x ?x-15
the rule for the signs is the same as in the simple case, put the first sign before the first x term. If the second sign is + put the same sign in both, if its – put the opposite sign before the second x term
8x2-?x+?x-15
we also use the same rule for the to coefficients of x, they must add or subtract to make the middle number(14) but they must times to make the end number times the first(15×8=120)
8x2-20x+6x-15
we then take out the common factor of the first 2 terms(4x)
4x(2x-5) + 6x -15
we use the bracket(2x-5) as the common factor for the second 2 terms and find what we need to multiply by(3)
4x(2x-5)+3(2x-5)
we then take the 2 numbers in front of the brackets(4x and 3) as our second bracket
(4x+3)(2x-5)
and there we have a fully factorized quadratic

Quadratic Formula

February 24, 2008 9 comments

The quadratic formula is a quick(unless you can factorise) way of solving quadratic equations. You basically take the coefficient’s of x, x2 and numbers, put then in the formula, work out the two answers and have your 2 solutions of x. And if that wasn’t easy enough written a console program in c++ that will solve them for you(and gives dodgy answers for complex solution ie) imaginary answers when there are no real roots.

so for the general quadratic
ax2+bx+c=0

qdrtc2.gif


So what do you do, well enter a, b and c from the general equation into the formula, work out the answers and they are your solutions.

Whats the plus/minus thingy. You might be wondering what the thing is after the -b, well its a plus or minus sign. Because when you work out a square root it can have to answers, eg)root 9 = +3 and -3, he formula takes this into account by saying you must use the plus and the minus answers. This therefore means you will be 2 solutions to the quadratic, which makes sense as the graph is a curve and it therefore must cut the x-axis twice.

When the root part is negative(before you find the root) there are no real roots, only complex ones. This means that the curve comes down above the x-axis and doesn’t cut it. If you work out the complex roots, using i for root(-1), your pair of answers will be a conjugate pair.

David Woodfords Quadratic Calculator
This is a console program that i wrote in c++ that will solve a quadratic for you. Just download the file and run it, follow the instructions and it will output the 2 answers for you. I tried to make it work for complex roots as well but somewhere in the decimal data types Ive messed up so only the second parts of the complex answers are correct – though im sure working out -b/2a isn’t too hard for you.
Download quadratic calculator


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