4 Solution by formula

When it is difficult to factorise a quadratic equation, it may be possible to solve it using a formula which is used to calculate the roots. The formula is obtained by completing the square in the general quadratic a x 2 + b x + c . We proceed by removing the coefficient of a :

a x 2 + b x + c = a { x 2 + b a x + c a } = a { ( x + b 2 a ) 2 + c a − b 2 4 a 2 }

Thus the solution of a x 2 + b x + c = 0 is the same as the solution to

( x + b 2 a ) 2 + c a − b 2 4 a 2 = 0

So, solving: ( x + b 2 a ) 2 = − c a + b 2 4 a 2 which leads to x = − b 2 a ± − c a + b 2 4 a 2

Simplifying this expression further we obtain the important result:

Key Point 5

Quadratic Formula

If a x 2 + b x + c = 0 , a ≠ 0 then the two solutions (roots) are

x = − b − b 2 − 4 a c 2 a and x = − b + b 2 − 4 a c 2 a

To apply the formula to a specific quadratic equation it is necessary to identify carefully the values of a , b and c , paying particular attention to the signs of these numbers. Substitution of these values into the formula then gives the desired solutions.

Note that if the quantity b 2 − 4 a c (called the discriminant ) is a positive number we can take its square root and the formula will produce two values known as distinct real roots . If b 2 − 4 a c = 0 there will be one value only known as a repeated root or double root . The value of this root is x = − b 2 a . Finally if b 2 − 4 a c is negative we say the equation possesses complex roots . These require special treatment and are described in HELM booklet  10.

Key Point 6

When finding roots of the quadratic equation a x 2 + b x + c = 0 first calculate the discrinimant

b 2 − 4 a c

∙ If b 2 − 4 a c > 0 the quadratic has two real distinct roots

∙ If b 2 − 4 a c = 0 the quadratic has two real and equal roots

∙ If b 2 − 4 a c < 0 the quadratic has no real roots: there are two complex roots

Example 20

Compare each given equation with the standard form a x 2 + b x + c = 0 and identify a , b and c . Calculate b 2 − 4 a c in each case and use this information to state the nature of the roots.

1.   3 x 2 + 2 x − 7 = 0 2.   3 x 2 + 2 x + 7 = 0
3.    3 x 2 − 2 x + 7 = 0 4.   x 2 + x + 2 = 0
5.   − x 2 + 3 x − 1 2 = 0 6.   5 x 2 − 3 = 0
7.   x 2 − 2 x + 1 = 0 8.   2 p 2 − 4 p = 0
9.   − p 2 + 4 p − 4 = 0
Solution
  1. a = 3 , b = 2 , c = − 7 . So b 2 − 4 a c = ( 2 ) 2 − 4 ( 3 ) ( − 7 ) = 88 .

    The roots are real and distinct.
  2. a = 3 , b = 2 , c = 7 . So b 2 − 4 a c = ( 2 ) 2 − 4 ( 3 ) ( 7 ) = − 80 .

    The roots are complex.
  3. a = 3 , b = − 2 , c = 7 . So b 2 − 4 a c = ( − 2 ) 2 − 4 ( 3 ) ( 7 ) = − 80 .

    The roots are complex.
  4. a = 1 , b = 1 , c = 2 . So b 2 − 4 a c = 1 2 − 4 ( 1 ) ( 2 ) = − 7 .

    The roots are complex.
  5. a = − 1 , b = 3 , c = − 1 2 . So b 2 − 4 a c = 3 2 − 4 ( − 1 ) ( − 1 2 ) = 7 .

    The roots are real and distinct.
  6. a = 5 , b = 0 , c = − 3 . So b 2 − 4 a c = 0 − 4 ( 5 ) ( − 3 ) = 60 .

    The roots are real and distinct.
  7. a = 1 , b = − 2 , c = 1 . So b 2 − 4 a c = ( − 2 ) 2 − 4 ( 1 ) ( 1 ) = 0 .

    The roots are real and equal.
  8. a = 2 , b = − 4 , c = 0 . So b 2 − 4 a c = ( − 4 ) 2 − 4 ( 2 ) ( 0 ) = 16

    The roots are real and distinct.
  9. a = − 1 , b = 4 , c = − 4 . So b 2 − 4 a c = ( − 4 ) 2 − 4 ( − 1 ) ( − 4 ) = 0

    The roots are real and equal.
Example 21

Solve the quadratic equation 2 x 2 + 3 x − 6 = 0 using the formula.

Solution

We compare the given equation with the standard form a x 2 + b x + c = 0 in order to identify a , b and c . We see that here a = 2 , b = 3 and c = − 6 . Note particularly the sign of c . Substituting these values into the formula we find

x = − b ± b 2 − 4 a c 2 a = − 3 ± 3 2 − 4 ( 2 ) ( − 6 ) ( 2 ) ( 2 ) = − 3 ± 9 + 48 4 = − 3 ± 7.5498 4

Hence, to 4 d.p., the two roots are x = 1.1375 , if the positive sign is taken and x = − 2.6375 if the negative sign is taken. However, it is often sufficient to leave the solution in the so-called surd form x = − 3 ± 57 4 , which is exact.

Task!

Solve the equation 3 x 2 − x − 6 = 0 using the quadratic formula.

First identify a , b and c :

a = 3 , b = − 1 , c = − 6

Substitute these values into the formula and simplify:

− ( − 1 ) ± ( − 1 ) 2 − ( 4 ) ( 3 ) ( − 6 ) ( 2 ) ( 3 ) = 1 ± 73 6

Finally, calculate the values of x to 4 d.p.:

1.5907 , − 1.2573