The negative solution is x = –1.
Solution:
Let the number be x.
Square of a number = x²
3 times a number = 3x
Difference of square of a number and 4 = 3 times that number
[tex]\Rightarrow x^2-4=3x[/tex]
Subtract 3x from both sides of the equation, we get
[tex]\Rightarrow x^2-3x-4=0[/tex]
–3x can be written as x – 4x.
[tex]\Rightarrow x^2+x-4x-4=0[/tex]
[tex]\Rightarrow( x^2+x)+(-4x-4)=0[/tex]
1st bracket have common term x and bracket have –4 common term.
[tex]\Rightarrow x( x+1)-4(x+1)=0[/tex]
Take common term x + 1 outside.
[tex](x+1)(x-4)=0[/tex]
By zero factor principle, if AB = 0 then A = 0 or B = 0.
x + 1 = 0, x – 4 = 0
x = –1, x = 4
The negative solution is x = –1.
