The formula given below is particularly useful for quadratics which cannot be factorised. To prove this important result requires some quite complex analysis, using a technique called completing the square, which is the subject of Section F4.3.
Theorem
The solutions of the quadratic equation
ax2 + bx + c = 0
are given by
Proof
The equation ax2 + bx + c = 0 is first divided by the non-zero constant, a, giving
| x2 + x + | = 0 | ||
| Note that | |||
| ⎛ ⎝x + ⎞ ⎠2 |
= ⎛ ⎝x + ⎞ ⎠ ⎛ ⎝x + ⎞ ⎠ | ||
| = x2 + + +
⎛ ⎝⎞ ⎠2 |
(expanding) | ||
| = x2 + +
⎛ ⎝⎞ ⎠2 | (adding like terms) | ||
| = x2 + +
⎛ ⎝⎞ ⎠2 | (simplifying) |
The first two terms are identical to the first two terms in our equation, so you can re-write the equation as
| ⎛ ⎝x + ⎞ ⎠2 − ⎛ ⎝⎞ ⎠2 + |
= 0 | |
| ⎛ ⎝x + ⎞ ⎠2 |
= ⎛ ⎝⎞ ⎠2 − | |
| = − | ||
| i.e. | ⎛ ⎝x + ⎞ ⎠2 |
= |
Taking the square root of both sides of the equation gives
| x + | = ± | |
| = | ||
| Hence | ||
| x | = − ± | |
| or | x | = |
as required.
Worked Examples
Solve
x2 + 6x − 8 = 0
giving the solution correct to 2 decimal places.
Here a = 1, b = 6 and c = −8. These values can be substituted into
| x | = | ||
| to give | |||
| x | = | ||
| = | |||
| = | or | ||
| = 1.12 | or −7.12 (to 2 d.p.) | ||
Solve the quadratic equation
4x2 − 12x + 9 = 0 .
Here a = 4, b = −12 and c = 9. Substituting the values into
| x | = | ||
| gives | |||
| x | = | ||
| = | |||
| = | |||
| = ⎛ ⎝= ⎞ ⎠ | |||
| = 1.5 |
Solve the quadratic equation
x2 + x + 5 = 0
Here a = 1, b = 1 and c = 5. Substituting the values into the formula gives
| x | = | |
| = | ||
| = |
As it is not possible to find , this equation has no solutions.
These three examples illustrate that a quadratic equation can have 2, 1 or 0 solutions.
The graphs below illustrate these graphically and show how the number of solutions
depends on the sign of
Exercises
