Question

how many solutions will each equation have?x²+4x+4=0 →x²-2x+1=0 →x²+6x+5=0 →x²-8x+16=0 →x²-4x+11=0 →x²-12x+13=0 →

Answer 1

Here, we want to get the number of solutions that each of the trinomial will have

To get this, we will have to use the term called the 'discriminant'

Mathematically, we have this as;

`D=b^2-4ac`

The value of the discriminant will dictate the number of solutions

At any point in time, a represents the leading coefficient which is also the coefficient of the term x^2

b is the coefficient of x

c is the last term

The value of D can either be positive, negative or zero

When positive, we have two real solutions

When negative, two imaginary solutions (we can say no solution at this level)

when zero, we have 1 real solution

a) a = 1 , b = 4 and c = 4

`D=4^2-4(1)(4)\text{ = 0}`

There is 1 real solution

b) a = 1, b = -2 and c = 1

`\begin{gathered} D=(-2)^2-4(1)(1) \\ D\text{ = 4-4 = 0} \end{gathered}`

There is 1 real solution

c) a = 1 , b = 6 and c = 5

`\begin{gathered} D=6^2-4(1)(5) \\ D\text{ = 36-20} \\ D\text{ = 16} \end{gathered}`

There are 2 real solutions

d) a = 1 , b = -8 and c = 16

`\begin{gathered} D=(-8)^2-4(1)(16) \\ D\text{ = 64-64 = 0} \end{gathered}`

There is 1 real solution

e) a = 1 , b = -4 and c = 11

`\begin{gathered} D=(-4)^2-4(1)(11) \\ D\text{ = 16-44} \\ D\text{ = -28} \end{gathered}`

There are no real solutions (however, there are 2 complex or imaginary solutions)

f) a = 1, b = -2 and c = 13

`\begin{gathered} D=(-2)^2-4(1)(13) \\ D\text{ = 4-52} \\ D\text{ = -48} \end{gathered}`

There are no real solutions (however, there are 2 complex or imaginary roots)