Answer:
The positive integer is 3.
Step-by-step explanation:
Let 'x' be the integer
Given that five times of a positive integer is less than twice its square by 3.
so
5 times of x = [tex]5x[/tex]
3 less than twice of the square of x = [tex]2x^{2} - 3[/tex]
so the equation becomes
[tex]5x = 2x^{2} - 3[/tex]
[tex]2x^2-5x-3=0[/tex]
Factorise
[tex]\left(2x+1\right)\left(x-3\right)=0[/tex]
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
[tex]2x+1=0\quad \mathrm{or}\quad \:x-3=0[/tex]
solving
[tex]2x+1=0[/tex]
Subtract 1 from both sides
[tex]2x+1-1=0-1[/tex]
Simplify
[tex]2x=-1[/tex]
Divide both sides by 2
[tex]\frac{2x}{2}=\frac{-1}{2}[/tex]
Simplify
[tex]x=-\frac{1}{2}[/tex]
Also solving
[tex]x-3=0[/tex]
Add 3 to both sides
[tex]x-3+3=0+3[/tex]
simplify
[tex]x=3[/tex]
Thus, we got integers:
x = 3 or x = -1/2
Therefore, the positive integer is 3.
