Question

Five times of a positive integer is less than twice its square by 3. Find the integer.
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Answer 1

Answer:

-1/2 or 3

Step-by-step explanation:

let the integer be x.

therefore, the equation will be

2x^2-5x-3=0

2x^2-6x+x-3=0 (middle term breaking)

2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x=3 or -1/2

Answer 2

$\Large \fbox \red {Question \: :}$

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

$\Large \fbox \red {Answer \: :}$

Let the integer be x.

• Five times of a positive integer = 5x
• Twice its square = 2x²
• Less by = 3

So, our equation is :

2x² - 5x = 3

2x² - 5x - 3 = 0 (Using splitting method)

2x² + x - 6x - 3

x (2x + 1) - 3 (2x + 1)

= (2x + 1) (x - 3)

★_______________★

• 2x + 1 = 0
• 2x = - 1
• x = - 1/2

★_______________ ★

• x - 3 = 0
• x = 3

★_______________★

$\implies$ So, the value of integer should be -½ or 3.

____________________&

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