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Degree: Chemical Engineering w/ industrial experience (Masters) - Manchester University
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
Suppose you have a quadratic equation: x2 + 5x + 6
You are told it has TWO REAL ROOTS and you are to find them.
The factorised quadratic with two roots will look like this:
Where a and b are real numbers (any positive or negative number really)
As this is just another form of the original equation we can say they are equal:
(x+a)(x+b) = x2 + 5x + 6
If you mulitply out the brackets using the FOIL (First, Outside, Inside, Last) rule you get:
x2 + xa + xb + ab = x2 + 5x + 6
It's clear the x2 terms cancel out, and if we equate the x terms and the number terms, we are left with
xa + xb = 5x and ab= 6
x(a+b) = 5x a x b = 6
a + b = 5
So now we must use this information to find a and b.
The factors of 6 are:
6 and 1
3 and 2
Of those factors, the pair which adds to 5 are 3 and 2.
so a = 3 and b = 2
Now we must check the signs of the factorised equation to check when we multiply ot the bracket we get the original equation again....
(x+3)(x+2) = x2 + 5x + 6 - CORRECT
Now we use this to find the roots, i.e. the x coordinates were y = 0
so therefore we make our factorised quadratic equal to 0
If two things multiplied together = 0, then at least one of them must equal 0...
x+3 = 0 ==> x = -3
x+2 = 0 ==> x = -2
We can check the roots are corrects by replacing a and b with the x terms in the original equation and it should equal 0 for both a and b.
We now have our roots...
x = -3
and x = -2see more