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a[sub]n[/sub] = 2n[sup]2[/sup] - 4n + 7
n>1
-
4
Ea[sub]i[/sub] 5+7+13+9=34
i=1
a[sub]4[/sub] = 2(4)[sup]2[/sup] - 4(4) + 7 = 32 - 23 = 9
Last edited:
a[sub]n[/sub] = 2n[sup]2[/sup] - 4n + 7
n>1
-
4
Ea[sub]i[/sub] 5+7+13+9=34
i=1
a[sub]4[/sub] = 2(4)[sup]2[/sup] - 4(4) + 7 = 32 - 23 = 9How about this
a[sub]n[/sub] = 2n[sup]2[/sup] - 4n + 7
n>1
-
4
Ea[sub]i[/sub] 5+7+13+23=48
i=1
a[sub]4[/sub] = 2(4)[sup]2[/sup] - 4(4) + 7 = 32 - 16 + 7 = 23The equation 2n² - 4n + 7 = 34 is used to solve for the value of n in a given quadratic equation. It can also be used to find the x-intercepts of a parabola.
To solve for n, you must first rearrange the equation to the standard form of a quadratic equation, which is ax² + bx + c = 0. Then, you can use the quadratic formula or factor the equation to find the value(s) of n.
When the equation specifies n > 1, it means that the solution for n must be greater than 1. This is a restriction placed on the possible values for n in order to make the equation meaningful.
Yes, this equation can also be solved by factoring. However, the quadratic formula is a more reliable method that can be used for any quadratic equation, regardless of its complexity.
Yes, this equation can have up to two solutions for n. This is because it is a quadratic equation, which means it has a degree of 2 and can have a maximum of two roots.