Question

Five times of a positive integer is less than twice its square by 3. Find the integer.

Answer 1

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 = $5x$

3 less than twice of the square of x = $2x^{2} - 3$

so the equation becomes

$5x = 2x^{2} - 3$

$2x^2-5x-3=0$

Factorise

$\left(2x+1\right)\left(x-3\right)=0$

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$2x+1=0\quad \mathrm{or}\quad \:x-3=0$

solving

$2x+1=0$

Subtract 1 from both sides

$2x+1-1=0-1$

Simplify

$2x=-1$

Divide both sides by 2

$\frac{2x}{2}=\frac{-1}{2}$

Simplify

$x=-\frac{1}{2}$

Also solving

$x-3=0$

Add 3 to both sides

$x-3+3=0+3$

simplify

$x=3$

Thus, we got integers:

x = 3 or x = -1/2

• As x can not be a negative integer according to the question.

Therefore, the positive integer is 3.