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Answer :
Sure! Let's solve the problem step-by-step.
The problem involves finding two consecutive odd integers whose product is less than 76. The first odd integer is denoted by [tex]\( n \)[/tex].
Odd integers increase by 2, so the two consecutive odd integers can be represented as [tex]\( n \)[/tex] and [tex]\( n + 2 \)[/tex].
The inequality we are dealing with is:
[tex]\[ n(n + 2) < 76 \][/tex]
To find the range of values for [tex]\( n \)[/tex], we will solve this inequality.
1. Set up the inequality:
[tex]\[ n(n + 2) < 76 \][/tex]
Expanding this, we have:
[tex]\[ n^2 + 2n < 76 \][/tex]
2. Rearrange the inequality:
Subtract 76 from both sides to form a quadratic inequality:
[tex]\[ n^2 + 2n - 76 < 0 \][/tex]
3. Solve the quadratic inequality:
To solve the quadratic inequality, we first determine the roots of the quadratic equation:
[tex]\[ n^2 + 2n - 76 = 0 \][/tex]
The roots are found using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -76 \)[/tex].
[tex]\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-76)}}{2 \cdot 1} \][/tex]
[tex]\[ n = \frac{-2 \pm \sqrt{4 + 304}}{2} \][/tex]
[tex]\[ n = \frac{-2 \pm \sqrt{308}}{2} \][/tex]
[tex]\[ n = \frac{-2 \pm \sqrt{4 \cdot 77}}{2} \][/tex]
[tex]\[ n = \frac{-2 \pm 2\sqrt{77}}{2} \][/tex]
[tex]\[ n = -1 \pm \sqrt{77} \][/tex]
Therefore, the roots are [tex]\( n = -1 + \sqrt{77} \)[/tex] and [tex]\( n = -1 - \sqrt{77} \)[/tex].
4. Determine the intervals:
Since we're looking for the values of [tex]\( n \)[/tex] that make the quadratic expression less than 0, we examine the number line and determine that [tex]\( n \)[/tex] is between the two roots.
Thus, the solution for [tex]\( n \)[/tex] is:
[tex]\(-1 - \sqrt{77} < n < -1 + \sqrt{77}\)[/tex]
This interval represents the values for [tex]\( n \)[/tex] where the product of the two consecutive odd integers is less than 76.
The problem involves finding two consecutive odd integers whose product is less than 76. The first odd integer is denoted by [tex]\( n \)[/tex].
Odd integers increase by 2, so the two consecutive odd integers can be represented as [tex]\( n \)[/tex] and [tex]\( n + 2 \)[/tex].
The inequality we are dealing with is:
[tex]\[ n(n + 2) < 76 \][/tex]
To find the range of values for [tex]\( n \)[/tex], we will solve this inequality.
1. Set up the inequality:
[tex]\[ n(n + 2) < 76 \][/tex]
Expanding this, we have:
[tex]\[ n^2 + 2n < 76 \][/tex]
2. Rearrange the inequality:
Subtract 76 from both sides to form a quadratic inequality:
[tex]\[ n^2 + 2n - 76 < 0 \][/tex]
3. Solve the quadratic inequality:
To solve the quadratic inequality, we first determine the roots of the quadratic equation:
[tex]\[ n^2 + 2n - 76 = 0 \][/tex]
The roots are found using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -76 \)[/tex].
[tex]\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-76)}}{2 \cdot 1} \][/tex]
[tex]\[ n = \frac{-2 \pm \sqrt{4 + 304}}{2} \][/tex]
[tex]\[ n = \frac{-2 \pm \sqrt{308}}{2} \][/tex]
[tex]\[ n = \frac{-2 \pm \sqrt{4 \cdot 77}}{2} \][/tex]
[tex]\[ n = \frac{-2 \pm 2\sqrt{77}}{2} \][/tex]
[tex]\[ n = -1 \pm \sqrt{77} \][/tex]
Therefore, the roots are [tex]\( n = -1 + \sqrt{77} \)[/tex] and [tex]\( n = -1 - \sqrt{77} \)[/tex].
4. Determine the intervals:
Since we're looking for the values of [tex]\( n \)[/tex] that make the quadratic expression less than 0, we examine the number line and determine that [tex]\( n \)[/tex] is between the two roots.
Thus, the solution for [tex]\( n \)[/tex] is:
[tex]\(-1 - \sqrt{77} < n < -1 + \sqrt{77}\)[/tex]
This interval represents the values for [tex]\( n \)[/tex] where the product of the two consecutive odd integers is less than 76.
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