Solving Quadratics

There are 2 ways to Solve quadratics
1 - By factoring
2 - Using the quadratic formula.

Solving Quadratics by Factoring:

The main difference between solving quadratics on the Higher Level course is that you will often have a number in front of the x² term, and the quadratic won't always be presented as a nice tidy x² + 4x + 4 = 0 format.

Dealing with the number in front of the x² term:
Looking at q 13
7x² + 29x + 4 = 0
As 7 is prime, the two brackets will be of the form
(7x .... )(x .....)=0
The numbers that multiply to make 4 are 4 × 1 and 2 × 2.
Trying e.g.
??? (7x + 2)(x + 2)= 0
If we multiply the inside pair and outside pair to check the x term we get
inside pair = 2x
outside pair = 14x
Total = 16x ≠ 29x
Trying again
(7x + 1)(x +4) = 0
Gives x + 28x = 29x so these factors are correct.
Factors are (7x + 1)(x +4) = 0
Roots are
7x + 1 = 0
x = -1/7
x+4=0
x = -4
Ans: Roots of equation are x = -1/7 or -4

With question 16
x(x-4) = 21
you have to tidy it up first
x² - 4x = 21
x² - 4x -21 = 0 (Note the =0 part is critical, because we will be exploiting the zero law to solve the factored equation).
Then factor as usual.

With questions 41 and 44 there is a lot more tidying up to do. Get rid of brackets and then put each equation in the form ....x² ....x ... number = 0

While you are looking at this topic, note the questions in the next section. There is no sign of a quadratic in e.g. question 5 but when you multiply accross by x to get rid of the fraction, you end up with x(x) = x²

Solving Quadratics Using the Quadratic Formula
If the question asks for the solution correct to 1 or 2 etc places of decimals, you need to use the quadratic formula.
Note that all the tidying up stuff mentioned above still applies, so you need to make sure the equation is in the form ax² + bx + c = 0

Looking at p35 q 13 you have
4x² + x -1 = 0
Which means that
a = 4, b = 1 (because x is the same as 1x) and c = -1 (note the sign).
Plugging numbers into the formula you get x =
-1 ± √( 1² - 4(4)(-1))
----------------------
 2(4)
= -1 ± √17
 --------
  8

The results are (-1 + 4.123)/8 = 0.3903 ≈ 0.39
  or (-1 - 4.123)/8 = -0.6403 ≈-0.64

Checking your answers in quadratics
Checking your answer is always a good idea, especially if you have time in an exam. Sometimes you may be asked to verify your answer. This means taking the result you got and plugging it back in to the original equation for x.

That means that in the examples above:

• if you substitute x = -4 into 7x² + 29x + 4 you should get zero.
• if you substitute x = 0.39 into 4x² + x -1 you should get very close to zero.

Can you explain why the answer in the second case is very close and not exactly = 0?