basic_engineering_mathematics0

Quadratic equations 71
In Problems 13 to 18, determine the quadratic equations in 3. Rearrange the equations so that the x 2 and x terms are on one
i.e. x 2 + 5 2 x − 3 2 = 0 x 2 + 9 2 x + 4 = 0
x whose roots are
13. 3 and 1 14. 2 and −5 15. −1 and −4
side of the equals sign and the constant is on the other side.
Hence
16. 2 1 2 and − 1 2
17. 6 and −6 18. 2.4 and −0.7
x 2 + 5 2 x = 3 2
4. Add to both sides of the equation (half the coefficient of x) 2 .In
this case the coefficient of x is
10.3 Solution of quadratic equations by
. Half the coefficient squared
( ) 2
‘completing the square’
5 2
is therefore . Thus
4
An expression such as x 2 or (x + 2) 2 or (x − 3) 2 is called a perfect
square.
If x 2 = 3 then x =± √ x 2 + 5 ( ) 5 2
2 x + = 3 ( ) 5 2
4 2 + 4
3
If (x + 2) 2 = 5 then x + 2 =± √ 5 and x =−2 ± √ 5
The LHS is now a perfect square, i.e.
If (x − 3) 2 = 8 then x − 3 =± √ 8 and x = 3 ± √ (
8
x + 5 ) 2
= 3 ( ) 5 2
Hence if a quadratic equation can be rearranged so that one side of
4 2 + 4
the equation is a perfect square and the other side of the equation 5. Evaluate the RHS. Thus
is a number, then the solution of the equation is readily obtained
(
by taking the square roots of each side as in the above examples.
x + 5 ) 2
= 3 The process of rearranging one side of a quadratic equation into a
4 2 + 25 24 + 25
= = 49
16 16 16
perfect square before solving is called ‘completing the square’.
6. Taking the square root of both sides of the equation
(x + a) 2 = x 2 + 2ax + a 2
Thus in order to make the quadratic expression x 2 + 2ax into a
(remembering that the square root of a number gives a
± answer). Thus
perfect square it is necessary to add (half the coefficient of x)
( ) 2
√
2a
2 (
i.e. or a 2
x + 5 )
√
2 ( ) 49
=
2
4 16
For example, x 2 + 3x becomes a perfect square by adding
( ) 3 2
, i.e.
2
i.e. x + 5 4 =±7 4
( ) 3 2 (
x 2 + 3x + = x + 3 ) 2
7. Solve the simple equation. Thus
2
2
x =− 5 The method is demonstrated in the following worked problems.
4 ± 7 4
i.e.
Problem 6. Solve 2x 2 + 5x = 3 by ‘completing the
x =− 5 4 + 7 4 = 2 4 = 1 2
square’.
and x =− 5 4 − 7 4 =−12 4 =−3
The procedure is as follows:
Hence x = 1 2
1. Rearrange the equation so that all terms are on the same side of 2x 2 + 5x = 3
or −3 are the roots of the equation
the equals sign (and the coefficient of the x 2 term is positive).
Hence 2x 2 + 5x − 3 = 0
Problem 7. Solve 2x 2 + 9x + 8 = 0, correct to 3 significant
2. Make the coefficient of the x 2 term unity. In this case this is
figures, by ‘completing the square’.
achieved by dividing throughout by 2. Hence
2x 2
2 + 5x
2 − 3 2 = 0
Making the coefficient of x 2 unity gives: