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NC1/ND1 Mathematics for Engineering
Quadratic Equations
A quadratic equation has no term with a power greater than the square e.g x2 - 5x + 6
This can be seen as the solution to the multiplication of the binomial expressions shown below:-
(x - 3)(x - 2)
The reversal of binomial multiplication is very useful and can be put to good effect in the solution
of equations. This reversal is known as factorising quadratic expressions.
The usual method of factorising is to look at all the factors of the first and last terms,
Then use a system of trial to find the combination which will provide the middle term.
e.g. Factorise 3x2 + 7x –6
Factors of 3x2
3x and x
Factors of -6
+1 and -6
-1 and +6
+2 and -3
-2 and +3
Then use trial and error e.g. (3x
(3x
(3x
(3x
+ 1)(x - 1)(x +
+ 2)(x - 2)(x +
6)
6)
3)
3)
=
=
=
=
3x2
3x2
3x2
3x2
- 17x - 6
+ 17x - 6
- 7x - 6
+ 7x - 6




Therefore the correct solution is (3x - 2)(x + 3)
Now if the equation had been given as 3x2 + 7x –6 = 0 then (3x - 2)(x + 3) = 0
Thus either (3x - 2) = 0 or (x + 3) = 0
This reasoning may be used to determine the roots of the equation (the algebraic values)
2
If 3x - 2 = 0 then 3x = +2 then x =
( 0.66 )
3
If x + 3 = 0 then x = -3
2
2
Check solution by substitution thus 3(-3)2 + 7(-3) –6 = 0 and 3( )2 + 7( ) - 6 = 0
3
3
This process may be reversed e.g.
Find the equation whose roots are (2,3)
In this case x = 2 or x = 3 Thus x - 2 = 0 or x - 3 = 0
Therefore:
( x – 2 )( x – 3) = 0 hence x2 - 5x + 6 = 0
x
b
b
2
2a
 4ac
when the standard form is
ax 2  bx  c  0
The most significant part of the formula is the expression
b 2  4ac
Note: If b2 > 4ac the roots can be found and will not be equal.
If b2 = 4ac the roots can be found and will be equal.
If b2 < 4ac the roots to negative numbers are involved:these are referred to as imaginary roots and will be investigated at a later stage.
Ref:-JRQuad1
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