Integer Square Condition Setup
Let the positive integer be represented by '$n$'. The problem states that the square of the integer ('$n^2$') is greater than 5 times the integer ('$5n$') by $-6$. This translates to the equation:
$n^2 = 5n + (-6)$
$n^2 = 5n - 6$
Solving the Quadratic Equation
To find the integer '$n$', we rearrange the equation into the standard quadratic form ($ax^2 + bx + c = 0$):
$n^2 - 5n + 6 = 0$
Now, we factor the quadratic expression:
$ (n - 2)(n - 3) = 0 $
This equation yields two possible integer solutions:
- $n - 2 = 0 \implies n = 2$
- $n - 3 = 0 \implies n = 3$
Identifying the Least Positive Integer
The possible integer values for '$n$' are 2 and 3. Both are positive integers.
The question asks for the least positive integer that satisfies the condition.
Comparing the solutions {2, 3}, the least value is 2.
Final Answer
The least positive integer satisfying the condition is 2.
Similar Questions
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- The possible value of x in the equation $\sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = \frac{25}{12}$ is:
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- Solve the following :
$\sqrt{7 + \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}}$ The sum of two positive numbers is 14 and their product is 45. The positive difference between them is:
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Important Questions from Quadratic Equation
For what values of k, the roots of 9x 2 + 8kx + 16 = 0 are real and equal?
- If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of \(\rm \left( 1+\frac{1}{x} \right) \) is equal to :
The nature of the roots of the equation 4x 2 - 2x - 3 = 0.
If the roots of the equation (q – r)x 2+ (r – p)x + (p – q) = 0 are equal, then which of the following is true?
Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is
